UNIVERSITY  OF  CALIFORNIA 
AT   LOS  ANGELES 


U.  or  & 


ELEMENTS  OF  GRAPHIC  STATICS. 


THE 


ELEMENTS  OF  GRAPHIC  STATICS 


for  &ttttients  of  Engineering 


L.    M.    HOSKINS 

PROFESSOR  OF  APPLIED  MECHANICS  IN  THE  LELAND  STANFORD 
JUNIOR  UNIVERSITY 


REVISED    EDITION 


Nefn 
THE  MACMILLAN  COMPANY 

LONDON :  MACMILLAN  &  CO..  LTD. 
1905 

All  rights  reserved 


COPYRIGHT,  1892, 
BY  M  ACM  ILL  AN  AND  CO. 


COPYRIGHT,  1899, 
BY  THE  MACMILLAN  COMPANY. 


Set  up  and  electrotyped  August,  1892.     New  edition,  revised, 
printed  from  new  plates  January,  1899;  March,  1903;  July,  1905. 


Korfoooo-  iBrcss 

J.  S.  Gushing  i  Co.  -  Berwick  St  Smith 
Norwood  Mass.  U.S.A. 


T  & 
2-70  v.otC, 


PREFACE   TO    REVISED   EDITION. 

THE  method  of  treatment  adopted  in  this  work  is  designed 
to  meet  the  needs  of  the  beginner.  To  this  end  the  endeavor 
has  been  made  to  secure  simplicity  of  presentation  without 
sacrifice  of  logical  rigor. 

In  scope,  the  work  has  been  planned  with  reference  to  the 
requirements  of  students  of  engineering.  This  limits  the  de- 
velopment of  the  general  theory  to  such  principles -and  methods 
as  are  practically  useful.  It  also  excludes  many  applications 
which,  though  leading  to  practical  results,  are  likely  to  prove 
useful  and  to  save  labor  only  in  the  hands  of  the  expert  in 
graphical  methods.  Graphic  Statics  is  treated  as  a  branch  of 
Mechanics  rather  than  of  Geometry,  and  those  beautiful  devel- 
opments whose  chief  interest  is  geometrical  have  not  been 
included. 

Although  graphical  methods  are  especially  useful  to  the 
structural  engineer,  it  is  believed  that  students  in  all  depart- 
ments of  engineering  will  find  it  profitable  to  become  familiar 
with  the  general  theory  of  complanar  forces  from  the  graphical 
side,  as  well  as  with  the  simpler  applications  to  the  determina- 
tion of  stresses  in  framed  structures,  and  of  centroids  and  mo- 
ments of  inertia  of  plane  areas.  The  application  to  trusses 
carrying  moving  loads  is  of  less  general  interest,  its  practical 
utility  being  limited  mainly  to  the  solution  of  problems  in 
bridge  design.  The  methods  developed  in  Chapter  VII  will 
therefore  be  of  practical  value  mainly  to  students  giving  special 
attention  to  this  branch  of  engineering. 

In  the  present  revised  edition  no  change  has  been  made  in 
general  plan,  and  few  changes  in  the  treatment  adopted,  except 

46325O 


vi  PREFACE   TO   THE   REVISED   EDITION. 

in  the  portions  relating  to  beams  and  trusses  carrying  moving 
loads  (Chapters  VI  and  VII).  These  portions  have  been 
wholly  re-written.  It  is  believed  that  a  substantial  improve- 
ment has  been  made  upon  the  methods  hitherto  used,  par- 
ticularly in  the  criterion  for  determining  the  position  of  a 
given  load-series  which  causes  maximum  stress  in  any  member 
of  a  truss.  The  improvement  consists  in  generalization,  which 
is  believed  to  be  gained  without  sacrifice  of  simplicity.  The 
graphical  method  of  applying  the  criterion  in  the  case  of 
trusses  with  parallel  chords  has  been  fully  treated  by  Pro- 
fessor H.  T.  Eddy.  The  method  here  given  applies  without 
the  restriction  to  parallel  chords.  The  algebraic  statement  of 
the  same  criterion,  as  given  in  Art.  152,  is  also  believed  to 
be  a  useful  generalization  of  the  methods  heretofore  used. 
Whether  the  algebraic  or  the  graphical  treatment  is  preferred, 
a  method  is  useful  in  proportion  to  its  generality,  provided  this 
does  not  involve  a  loss  of  simplicity.  There  is  a  decided  ad- 
vantage in  the  use  of  a  single  general  equation,  applicable  to 
any  member  of  any  truss,  instead  of  several  particular  equa- 
tions, each  applicable  to  a  special  member  or  to  a  special  form 
of  truss. 

STANFORD  UNIVERSITY,  CALIFORNIA, 
November,  1898. 


CONTENTS. 


PART   I.  — GENERAL   THEORY. 
CHAPTER  I.    DEFINITIONS. — CONCURRENT  FORCES. 

PAGE 

1 .  Preliminary  Definitions i 

2.  Composition  of  Concurrent  Forces       5 

3.  Equilibrium  of  Concurrent  Forces 6 

4.  Resolution  of  Concurrent  Forces 8 

CHAPTER  II     NON-CONCURRENT  FORCES. 

1.  Composition  of  Ncn-concurrent  Forces  Acting  on  the  Same  Rigid 

Body ii 

2.  Equilibrium  of  Non-concurrent  Forces 18 

3.  Resolution  into  Non-concurrent  Systems 28 

4.  Moments  of  Forces  and  of  Couples 30 

5.  Graphic  Determination  of  Moments 35 

6.  Summary  of  Conditions  of  Equilibrium 37 

CHAPTER  III.     INTERNAL  FORCES  AND  STRESSES. 

1.  External  and  Internal  Forces 40 

2.  External  and  Internal  Stresses 42 

3.  Determination  of  Internal  Stresses 46 


PART   II.  — STRESSES   IN   SIMPLE   STRUCTURES. 

CHAPTER  IV.     INTRODUCTORY. 
§  i .    Outline  of  Principles  and  Methods 


S3 


CONTENTS. 


CHAPTER  V.    ROOF  TRUSSES.  —  FRAMED  STRUCTURES  SUSTAINING 
STATIONARY  LOADS. 

PAGE 

1.  Loads  on  Roof  Trusses 58 

2.  Roof  Truss  with  Vertical  Loads 61 

3.  Stresses  Due  to  Wind  Pressure 67 

4.  Maximum  Stresses 71 

5.  Cases  Apparently  Indeterminate 74 

6.  Three-hinged  Arch 80 

7.  Counterbracing 88 

CHAPTER  VI.     SIMPLE  BEAMS. 

1.  General  Principles 94 

2.  Beam  Sustaining  Fixed  Loads 98 

3.  Beam  Sustaining  Moving  Loads 101 

CHAPTER  VII.    TRUSSES  SUSTAINING  MOVING  LOADS. 

1.  Bridge  Loads       112 

2.  Truss  Regarded  as  a  Beam 115 

3.  Truss  Sustaining  Any  Series  of  Moving  Loads 120 

4.  Truss  with  Subordinate  Bracing 137 

5.  Uniformly  Distributed  Moving  Load 145 


PART   III.  — CENTROIDS   AND   MOMENTS   OF   INERTIA. 
CHAPTER  VIII.    CENTROIDS. 

1.  Centroid  of  Parallel  Forces 152 

2.  Center  of  Gravity  —  Definitions  and  General  Principles  ....  157 

3.  Centroids  of  Lines  and  of  Areas 160 

CHAPTER  IX.    MOMENTS  OF  INERTIA. 

1.  Moments  of  Inertia  of  Forces 167 

2.  Moments  of  Inertia  of  Plane  Areas 177 

CHAPTER  X.    CURVES  OF  INERTIA. 

1.  General  Principles 187 

2.  Inertia-Ellipses  for  Systems  of  Forces 190 

3.  Inertia-Curves  for  Plane  Areas 195 


GRAPHIC    STATICS. 

PART   I. 
GENERAL    THEORY. 


CHAPTER   I.  — DEFINITIONS.     CONCURRENT 
FORCES. 

§  I.    Preliminary  Definitions. 

1.  Dynamics  treats  of  the  action  of  forces  upon  bodies.     Its 
two  main  branches  are  Statics  and  Kinetics. 

Statics  treats  of  the  action  of  forces  under  such  conditions 
that  no  change  of  motion  is  produced  in  the  bodies  acted  upon. 

Kinetics  treats  of  the  laws  governing  the  production  of  motion 
by  forces. 

2.  Graphic  Statics  has  for  its  object  the  deduction  of  the 
principles  of  statics,  and  the  solution  of  its  problems,  by  means 
of  geometrical  figures. 

3.  A  Force  is  that  which  tends  to  change  the  state  of  motion 
of  a  body.     We  conceive  of  a  force  as  a  push  or  a  pull  applied 
to  a  body  at  a  definite  point  and  in  a  definite  direction.     Such 
a  push  or  pull  tends  to  give  motion  to  the  body,  but  this  ten- 
dency may  be  neutralized  by  the  action  of  other  forces.     The 
effect  of  a  force  is  completely  determined  when  three  things 
are  given,  —  its  magnitude,  its  direction,  and  its  point  of  applica- 
tion.    The  line  parallel  to  the  direction  of  the  force  and  con- 
taining its  point  of  application,  is  called  its  line  of  action. 


GRAPHIC   STATICS. 


Every-  force  acting  upon  a  body  is  exerted  by  some  other  body. 
But  the  problems  of  statics  usually  concern  only  the  body  acted 
upon.  Hence,  frequently,  no  reference  is  made  to  the  bodies 
exerting  the  forces. 

4.  Unit  Force.  —  The  unit  force  is  a  force  of  arbitrarily  chosen 
magnitude,  in  terms  of  which  forces  are  expressed.     Several 
different  units  are  in  use.     The  one  employed  in  this  work  is 
the  pound,  which  will  now  be  denned. 

A  pound  force  is  a  force  equal  to  the  weight  of  a  pound  mass 
at  the  earth's  surface.  A  pound  mass  is  the  quantity  of  matter 
contained  in  a  certain  piece  of  platinum,  arbitrarily  chosen,  and 
established  as  the  standard  by  act  of  the  British  Parliament. 

The  pound  force,  as  thus  defined,  is  not  perfectly  definite, 
since  the  weight  of  any  given  mass  (that  is,  the  attraction  of 
the  earth  upon  it)  is  not  the  same  for  all  positions  on  the  earth's 
surface.  The  variations  are,  however,  unimportant  for  most  of 
the  requirements  of  the  engineer. 

In  its  fundamental  meaning,  the  word  "  pound  "  refers  to  the 
unit  mass,  and  it  is  unfortunate  that  it  is  also  applied  to  the  unit 
force.  The  usage  is,  however,  so  firmly  established  that  it  will 
be  here  followed. 

5.  Concurrent  and  Non-concurrent  Forces.  —  Forces  acting  on 
the  same  body  are  concurrent  when  they  have  the  same  point  of 
application.     When  applied  at  different  points  they  are   non- 
concurrent. 

6.  Complanar  Forces  are  those  whose  lines  of  action  are  in 
the  same  plane.     In  this  work,  only  complanar  systems  will  be 
considered  unless  otherwise  specified. 

.  7.  A  Couple  is  the  name  given  to  a  system  consisting  of  two 
forces,  equal  in  magnitude,  but  opposite  in  direction,  and  having 
different  lines  of  action.  The  perpendicular  distance  between 
the  two  lines  of  action  is  called  the  arm  of  the  couple. 


PRELIMINARY   DEFINITIONS. 

8.  Equivalent  Systems  of  Forces.  —  Two  system 
are  equivalent  when  either  may  be  substituted  for 
without  change  of  effect. 

9.  Resultant.  —  A  single  force  that  is  equivalent  to  a  given 
system  of  forces  is  called  the  resultant  of  that  system.     It  will 
be  shown  subsequently  that  a  system   of  forces  may  not   be 
equivalent  to  any  single  force.     When  such  is  the  case,  the 
simplest  system  equivalent  to  the  given  system  may  be  called 
its   resultant.     Any   forces    having    a    given    force    for   their 
resultant  are  called  components  of  that  force. 

10.  Composition  and  Resolution  of   Forces.  —  Having   given 
any  system  of  forces,  the  process  of  finding  an  equivalent  system 
is  called  the  composition  of  forces,  if  the  system  determined  con- 
tains fewer  forces  than  the  given  system.     If  the  reverse  is  the 
case,  the  process  is  called  the  resolution  of  forces. 

The  process  of  finding  the  resultant  of  any  given  forces  is 
the  most  important  case  of  composition ;  while  the  process  of 
finding  two  or  more  forces,  which  together  are  equivalent  to  a 
single  given  force,  is  the  most  common  case  of  resolution. 

11.  Representation  of  Forces  Graphically. — The  magnitude 
and  direction  of  a  force  can  both  be  represented  by  a  line ;  the 
length  of  the  line  representing  the  magnitude  of  the  force,  and 
its  direction  the  direction  of  the  force. 

In  order  that  the  length  of  a  line  may  represent  the  magni- 
tude of  a  force,  a  certain  length  must  be  chosen  to  denote  the 
unit  force.  Then  a  force  of  any  magnitude  will  be  represented 
by  a  length  which  contains  the  assumed  length  as  many  times 
as  the  magnitude  of  the  given  force  contains  that  of  the  unit 
force. 

In  order  that  the  direction  of  a  force  may  be  represented  by 
a  line,  there  must  be  some  means  of  distinguishing  between  the 
two  opposite  directions  along  the  line.  The  usual  method  is  to 
place  an  arrow-head  on  the  line,  pointing  in  the  direction  toward 


4  GRAPHIC   STATICS. 

which  the  force  acts.  If  the  line  is  designated  by  letters  placed 
at  its  extremities,  the  order  in  which  these  are  read  may  indicate 
the  direction  of  the  force;  Thus,  AB  and  BA  represent  two 
forces,  equal  in  magnitude  but  opposite  in  direction. 

The  line  of  action  of  a  force  can  also  be  represented  by  a  line 
drawn  on  the  paper. 

In  solving  problems  in  statics,  it  is  usually  convenient  to 
draw  two  separate  figures,  in  one  of  which  the  forces  are 
represented  in  magnitude  and  direction  only,  and  in  the  other 
in  line  of  action  only. 

These  two  species  of  diagrams  will  be  called  force  diagrams 
and  space  diagrams,  respectively. 

12.  Notation.  —  The  use  of  graphic  methods  is  much  facili- 
tated by  the  adoption  of  a  convenient  system  of  notation  in 
the  figures  drawn. 

There  will  generally  be  two  figures  (the  force  diagram  and 
the  space  diagram)  so  related  that  for  every  line  in  one  there 
is  a  corresponding  line  in  the  other. 

In  the  force  diagram  each  line  represents  a  force  in  magni- 
tude and  direction ;    in   the  space  diagram  the  corresponding 
line  represents  the  line  of  action  of  the  force.     These   lines 
will  usually  be  designated  in  the  following  manner :  The  line 
denoting  the  magnitude  and    direction  of 
the  force  will  be  marked  by  two   capital 
letters,  one  at  each  extremity  ;   while  the 
action-line   will    be   marked  by  the   corre- 
sponding small  letters,  one  being  placed  at 
Fig  x  each  side  of  the  line  designated.     Thus,  in 

Fig.  i,  AB  represents  a  force  in  magni- 
tude and  direction,  while  its  action-line  is  marked  by  the  letters 
ab,  placed  as  shown. 


COMPOSITION   OF   CONCURRENT   FORCES. ^j^^gT^ 

§  2.    Composition  of  Concurrent  Forces\ 

13.  Resultant  of  Two  Concurrent  Forces.  —  If  tw^Jftg^tt^-r 
rent  forces  are  represented  in  magnitude  and  direction  by  two 
lines  AB  and  BC,  their 

resultant  is  represented  .,  c  c/b 

in  magnitude  and  direc-  ^^-""^^ J  (S) 

tion  by  AC.  (Fig.  2.) 
Proofs  of  this  proposi- 
tion are  given  in  all  "B 

Eig.  & 

elementary  treatises    on 

mechanics,  and  the  demonstration  will  be  here  omitted.  The 
point  of  application  of  the  resultant  is  the  same  as  that  of  the 
given  forces.  Thus  if  O  (Fig.  2)  is  the  given  point  of  applica- 
tion, then  ab,  be,  and  ac,  drawn  parallel  to  AB,  BC,  and  AC 
respectively,  are  the  lines  of  action  of  the  two  given  forces  and 
their  resultant.  The  figure  marked  (A)  is  a  force  diagram,  and 
(B)  is  the  corresponding  space  diagram  (Art.  11). 

14.  Resultant  of  Any  Number  of  Concurrent  Forces.  —  If  any 
number  of  concurrent  forces  are  represented  in  magnitude  and 
direction  by  lines  AB,  BC,  CD,  .  .  .,  their  resultant  is  repre- 
sented in  magnitude  and  direction  by  the  line  AN,  where  N  is 
the  extremity  of  the  line  representing  the  last  of   the  given 
forces. 

This    proposition   follows    immediately   from    the    preceding 
one ;  for  the  resultant  of   the  forces  represented  by  AB  and 
BC   is    a    force   repre- 
sented by  AC;  the  re- 
sultant of  AC  and  CD 
is  AD,  and  so  on.     By 
continuing  the  process 

we  shall  arrive  at  the  ^~"~^E       Fis- 3 

result     stated.       It     is 

readily  seen  that  the  order  in  which  the  forces  are  taken  does 
not  affect  the  magnitude  or  direction  of  the  resultant  as  thus 


6  GRAPHIC   STATICS. 

determined.  The  point  of  application  of  the  resultant  is  the 
same  as  that  of  the  given  forces.  Figure  3  shows  the  force 
diagram  and  space  diagram  for  a  system  of  four  forces  repre- 
sented by  AB,  BC,  CD,  DE,  and  their  resultant  represented 
by  AE,  applied  at  the  point  O. 

From  the  above  construction  it  is  evident  that  every  system 
of  concurrent  forces  has  for  its  resultant  some  single  force 
(Art.  9) ;  though  in  particular  cases  its  magnitude  may  be  zero. 

15.  Force  Polygon. — The  figure  formed  by  drawing  in  suc- 
cession   lines   representing    in    magnitude   and    direction   any 

number  of  forces  is  called  a  force  polygon  for 
those  forces.  Thus  Fig.  4  is  a  force  polygon 
for  any  four  forces  represented  in  magnitude 
and  direction  by  the  lines  AB,  BC,  CD,  and 
DE,  whatever  their  lines  of  action.  It  may 
happen  that  the  point  E  coincides  with  A,  in 
which  case  the  polygon  is  said  to  be  closed.  It  is  evident  that 
the  order  in  which  the  forces  are  taken  does  not  affect  the 
relative  positions  of  the  initial  and  final  points. 

§  3.    Equilibrium  of  Concurrent  Forces. 

1 6.  Definition. — A  system    of    forces  acting  on  a  body  is 
in  equilibrium  if  the  motion  of  the  body  is  unchanged  by  its 
action. 

17.  Condition   of    Equilibrium.  —  In   order   that   no   motion 
may  result  from  the  action  of  any  system  of  concurrent  forces, 
the  magnitude  of  the  resultant  must  be  zero ;  and  conversely, 
if    the   magnitude   of    the   resultant    is   zero,    no    motion    can 
result.     But   (Arts.    14   and  15)  the  condition  that  the  result- 
ant  is   zero   is   identical   with   the    condition    that    the    force 
polygon  closes.     Hence,  the  following  proposition  : 

If  any  sy stein  of  concurrent  forces  is  in  equilibrium,  the  force 
polygon  for  the  system  must  close.  And  conversely,  If  the  force 


EQUILIBRIUM   OF   CONCURRENT   FORCES.  7 

polygon  is  closed  for  any  system  of  concurrent  forces,  the  system 
is  in  equilibrium. 

The   comparison   of   this  with  the   analytical   conditions   of 
equilibrium  is  given  in  Art.  22. 

1 8.  Method  of  Solving  Problems  in  Equilibrium.  —  If  a  sys- 
tem of  concurrent  forces  in  equilibrium  be  partially  unknown, 
we  may  in  certain  cases  determine  the  unknown  elements  by 
applying  the  principles  of  Art.  17. 

A  common  case  is  that  in  which  two  forces  are  unknown 
except  as  to  lines  of  action.  Thus,  suppose  a  system  of  five 
forces  in  equilibrium, 
three  being  fully 
known,  represented  in 
magnitude  and  direc- 
tion by  AB,  BC,  CD 
(Fig.  5),  and  in  lines 
of  action  by  ab,  be,  cd, 
while  concerning  the  other  two  we  know  only  their  lines  of 
action  de,  ea.  To  determine  these  two  in  magnitude  and 
direction,  it  is  necessary  only  to  complete  the  force  polygon  of 
which  ABCD  is  the  known  part.  The  remaining  sides  must  be 
parallel  respectively  to  de  and  ea.  From  D  draw  a  line  parallel 
to  de,  and  from  A  a  line  parallel  to  ea,  prolonging  them  till  they 
intersect  at  E.  Then  DE  and  EA  represent  the  required  forces 
in  magnitude  and  direction,  and  the  complete  force  polygon  is 
ABCDEA.  It  is  evident  that  ABODE' A  is  an  equally  legiti- 
mate form  of  the  force  polygon,  and  gives  the  same  result  for 
the  magnitude  and  direction  of  each  of  the  unknown  forces. 
This  problem  occurs  constantly  in  the  construction  of  stress 
diagrams  by  the  method  described  in  Part  II. 

The  student  will  find  little  difficulty  in  treating  other  prob- 
lems in  the  equilibrium  of  concurrent  forces. 

19.  Problems  in  Equilibrium.  —  (i)  A  particle  is  in  equilib- 
rium   under    the    action    of    five    forces,    three   of    which   are 


8  GRAPHIC   STATICS. 

completely  known,  while  of  the  remaining  two,  one  is  known 
in  direction  only,  and  the  other  in  magnitude  only.  To  deter- 
mine the  unknown  forces. 

(2)  Suppose   two   forces    known    in    magnitude    but    not    in 
direction,  the  remaining  forces  being  wholly  known. 

(3)  Suppose   one   force  wholly  unknown,  the   others  being 
known. 

§  4.    Resolution  of  Concurrent  Forces. 

20.  To  Resolve   a  Given  Force  into  Any  Number  of  Com- 
ponents having  the  same  point  of  application,  we  have  only  to 
draw  a  closed  polygon  of  which  one  side  shall  represent  the 
magnitude  and  direction  of  the  given  force  ;  then  the  remaining 
sides  will  represent,  in  magnitude  and  direction,  the  required 
components.     This  problem  is,  in  general,  indeterminate,  unless 
the  components  are  required  to  satisfy  certain  specified  con- 
ditions. 

[NOTE.  —  A  problem  is  said  to  be  indeterminate  if  its  conditions  can  be  satisfied 
in  an  infinite  number  of  ways.  It  is  determinate  if  it  admits  of  only  one  solution. 
Thus,  the  problem,  to  determine  the  values  of  x  and  y  which  shall  satisfy  the  equation 
x  +  y  =  10,  is  indeterminate;  while  the  problem,  given  2  x -\-  3=  7,  to  find  the 
value  of  x,  is  determinate.  The  case  in  which  a  finite  number  of  solutions  is  possible 
may  be  called  incompletely  determinate.  Thus,  the  problem,  given  x'2  +  x  —  6  —  o, 
to  find  x,  admits  of  two  solutions,  and  therefore  is  incompletely  determinate.  All 
these  classes  of  problems  may  be  met  with  in  statics.] 

21.  To  Resolve  a  Given  Force  into  Two  Components.  —  This 
problem  is  indeterminate  unless  additional  data  are  given.     For 
if  the  given  force  be  represented  in  magnitude  and  direction  by 
a  line,  any  two  lines  which  with  the  given  line  form  a  triangle 
may  represent  forces  which  are  together  equivalent  to  the  given 
force.    'But  an  infinite  number  of  such  triangles  may  be  drawn. 
The  solution  of  the  following  four  cases  of  this  problem  will 
form  exercises  for  the  student.     In  each  case  the  force  diagram 
and  space  diagram  should  be  completely  drawn,  and  the  student 
should  notice  whether  the   problem    is    determinate,   partially 
determinate,  or  indeterminate. 


RESOLUTION   OF   CONCURRENT   FORCES//  Q    „ 

(1)  Let  the  lines  of  action  of  the  required  comp^nejits  be 
given 

(2)  Let  the  two  components  be  given  in  magnitude  only. 

(3)  Let  the  line  of  action  of  one  component  and  the  magni- 
tude of  the  other  be  given. 

(4)  Let  the  magnitude  and  direction  of  one  component  be 
given. 

It  will  be  noticed  that  these  four  cases  correspond  to  four 
cases  of  the  solution  of  a  plane  triangle. 

22.  Resolved  Part  of  a  Force.  —  If  a  force  is  conceived  to  be 
replaced  by  two  components  at  right  angles  to  each  other,  each 
is  called  the  resolved  part,*  in  its  direction,  of  the  given  force. 

It  is  readily  seen  that  the  resolved  part  of  a  force  repre- 
sented by  AB  (Fig.  6)  in  the  direction   of  any  line   XX,   is 
represented  in  magnitude  and  direction 
by  A'B',  the  orthographic  projection  of 
AB  upon   XX.      It   follows    that   the 
resolved  part  (in  any  given  direction)  of 
the  resultant  of  any  concurrent  forces 
is  equal   to  the  algebraic  sum  of  the 

resolved  parts  of  its  components  in  that  direction  ;  signs  plus 
and  minus  being  given  to  the  resolved  parts  to  distinguish  the 
two  opposite  directions  which  they 
may  have.     Thus  (Fig.   7)  the   re- 
solved parts  of  the  forces  AB,  BC, 
CD,  in  a  direction  parallel  to  XX, 
are   A'B',    B'C,    CD';    and   their 
algebraic  sum  is  A'D',  which  is  the 
resolved  part  of  the  resultant  AD. 
If  the  resultant  is  zero,  D'  coincides  with  A' ;  hence, 

(i)  For  the  equilibrium  of  any  concurrent  forces,  the  sum 
of  their  resolved  parts  in  any  direction  must  be  zero. 


*  The  term  "  resolute  "  has  been  proposed  by  J.  B.  Lock  ("  Elementary  Statics  ")  to 
denote  what  is  here  denned  as  the  resolved  part  of  a  force. 


I0  GRAPHIC   STATICS. 

Again,  if  D'  coincides  with  A',  then  either  D  coincides 
with  ,4?.pr.,else  AD  is  perpendicular  to  XX ";  hence, 

(2)  Tf  the  sum  of  the  resolved  parts  of  any  concurrent 
forces  in  a  given  direction  is  zero,  their  resultant  (if  any) 
is  perpendicular  to  that  direction.  And  if  the  sum  of  the 
resolved  parts  is  zero  for  each  of  two  directions,  the  resultant 
is  zero,  and  the  system  is  in  equilibrium. 

Propositions  (i)  and  (2)  state  the  conditions  of  equilibrium 
for  concurrent  forces  usually  deduced  in  treatises  employing 
algebraic  methods. 


CHAPTER   II.  — NON-CONCURRENT   FORCES. 

§   i.   Composition  of  Non-concurrent  Forces  Acting  on  the  Same 
Rigid  Body. 

23.  Definition  of  Rigid  Body. — A  rigid  body  is  one  whose 
particles  do  not  change  their  positions  relative  to  each  other 
under  any  applied  forces.     No  known  body  is  perfectly  rigid, 
but  for  the  purposes  of  statics,  most  solid  bodies  may  be  con- 
sidered as  such  ;  and  any  body  which  has  assumed  a  form  of 
equilibrium   under   applied   forces,   may,  for  the   purposes   of 
statics,  be  treated  as  a  rigid  body  without  error. 

24.  Change  of  Point  of  Application.  —  The  effect  of  a  force 
upon  a  rigid  body  will  be  the  same,  at  whatever  point  in  its 
line  of  action  it  is  applied,  if  only  the  particle  upon  which  it 
acts  is  rigidly  connected  with  the  body. 

This  proposition  is  fundamental  to  the  development  of  the 
principles  of  statics,  and  is  amply  justified  by  experience.* 
In  applying  the  principle,  we  are  at  liberty  to  assume  a 
point  of  application  outside  the  actual  body,  the  latter  being 
ideally  extended  to  any  desired  limits. 

25.  Resultant   of   Two   Non-Parallel   Forces.  —  If    two  corn- 
planar  forces  are  not  parallel,  their  lines  of  action  must  inter- 
sect, and  the  point  of  intersection  may  be  taken  as  the  point 
of  application  of  each  force.     Hence,  they  may  be  treated  as 


*  This  proposition  may  be  proved  analytically  by  deducing  the  equations  of  motion 
of  a  rigid  body,  and  showing  that  the  effect  of  any  force  on  the  motion  of  the  body 
depends  only  upon  its  magnitude,  direction,  and  line  of  action.  But  such  a  proof  is,  of 
course,  outside  the  scope  of  this  work. 

II 


12 


GRAPHIC   STATICS. 


concurrent  forces,  and  their  resultant  may  be  determined  as 
in  Art.  13.  The  following  proposition  may  therefore  be 
stated : 

If  two  forces  acting  in  the  same  plane  on  a  rigid  body 
are  represented  in  magnitude  and  direction  by  AB  and  BC, 
their  resultant  is  represented  in  magnitude  and  direction  by 

AC,  and  its  line  of  action  passes  through  the  point  of  inter- 
section of  the  lines  of  action  of  the  given  forces.     Its  point 
of  application  may  be  any  point  of  this  line. 

It  may  happen  that  the  point  of  intersection  of  the  two 
given  lines  of  action  falls  outside  the  limits  available  for  the 
drawing.  In  such  a  case  it  will  be  most  convenient  to  find 
the  resultant  by  the  method  to  be  explained  in  Art.  27.  The 
same  remark  applies  to  the  case  of  two  parallel  forces. 

26.  Resultant  of  Any  Number  of  Non-concurrent  Forces  — 
First  Method.  —  The  method  of  the  preceding  article  may  be 
extended  to  the  determination  of  the  resultant  of  any  number 
of  forces  acting  on  the  same  rigid  body.  Let  AB,  BC,  CD,  DE 

(Fig.  8),  represent  in 
magnitude  and  direction 
four  forces,  and  let  ab, 
be,  cd,  de  represent  their 
lines  of  action.  To  find 
their  resultant,  we  may 
proceed  as  follows : 
The  resultant  of  AB 
and  BC  is  represented 
in  magnitude  and  direction  by  AC,  and  in  line  of  action  by  ac 
drawn  parallel  to  AC  through  the  point  of  intersection  of  ab 
and  be.  Combining  this  resultant  with  CD,  we  get  as  their 
resultant  a  force  represented  in  magnitude  and  direction  by 

AD,  and  in  line  of  action  by  ad  drawn  parallel  to  AD  through 
the  point  of  intersection  of  ac  and  cd.     This  is  evidently  the 
resultant  of  AB,  BC,  and  CD.     In  the  same  way,  this  resultant 


Fig.  8 


COMPOSITION   OF   NON-CONCURRENT   FORCES.  13 

combined  with  DE  gives  for  their  resultant  a  force  whose  mag- 
nitude and  direction  are  represented  by  AE,  and  whose  line  of 
action  is  ae,  parallel  to  AE  and  passing  through  the  point  in 
which  ad  intersects  de.  This  last  force  is  the  resultant  of  the 
four  given  forces. 

The  process  may  evidently  be  extended  to  the  case  of  any 
number  of  forces. 

As  in  the  case  discussed  in  the  preceding  article,  this  method 
will  become  inapplicable  or  inconvenient  if  any  of  the  points  of 
intersection  fall  outside  the  limits  available  for  the  drawing. 
For  this  reason  it  is  usually  most  convenient  to  employ  the 
method  described  in  Art.  27. 

The  student  should  bear  in  mind  that  the  length  and  direc- 
tion AE  and  the  line  ae  are  not  the  magnitude,  direction,  and 
line  of  action  of  any  actual  force  applied  to  the  body.  By  the 
resultant  is  meant  an  ideal  force,  which,  if  it  acted,  would 
produce  the  same  effect  upon  the  motion  of  the '  body  as  is 
produced  by  the  given  forces.  It  is  a  force  which  may  be 
conceived  to  replace  the  actual  forces,  and  may  be  assumed  to 
be  applied  to  any  particle  in  its  line  of  action,  provided  that 
particle  is  regarded  as  rigidly  connected  with  the  given  body. 
The  line  of  action  may  in  reality  fail  to  meet  the  given  body. 
(See  Art.  24.) 

27.  Resultant  of  Non-concurrent  Forces  —  Second  Method. — 
This  method  will  be  described  by  reference  to  an  example. 
Referring  to  Fig.  9,  let  AB,  BC,  CD,  DE  represent  in  magni- 
tude and  direction  four  forces  whose  lines  of  action  are  ab,  be, 
cd,  de  ;  and  let  it  be  required  to  find  their  resultant.  Draw  the 
force  polygon  ABCDE,  and  from  any  point  O  in  the  force 
diagram  draw  lines  OA,  OB,  OC,  OD,  OE.  These  lines  may 
represent,  in  magnitude  and  direction,  components  into  which 
the  given  forces  may  be  resolved.  Thus  AB  is  equivalent  to 
forces  represented  by  AO  and  OB  acting  in  any  lines  parallel 
to  A  0,  OB,  whose  point  of  intersection  falls  upon  ab ;  BC  is 


I4  GRAPHIC   STATICS. 

equivalent  to  forces  represented  by  BO,  OC,  acting  in  any  lines 
parallel  to  BO,  OC,  which  intersect  on  be ;  and  so  for  each  of 
the- -given  forces.  The  four  given  forces  may,  therefore,  be 
replaced  by  eight  forces  given  in  magnitude  and  direction  by 


AO,  OB,  BO,  OC,  CO,  OD,  DO,  OE,  with  proper  lines  of 
action.  Now,  it  is  possible  to  make  the  lines  of  action  of  the 
forces  represented  by  OB  and  BO  coincide ;  and  the  same  is 
true  of  each  pair  of  equal  and  opposite  forces,  OC,  CO ;  OD, 
DO.  To  accomplish  this,  let  AO,  OB  act  in  lines  ao,  ob,  inter- 
secting at  any  assumed  point  of  ab.  Prolong  ob  to  intersect 
be,  and  take  the  point  thus  determined  as  the  point  of  inter- 
section of  the  lines  of  action  of  BO,  OC ;  these  lines  are  then 
bo,  oc.  Similarly  prolong  oc  to  intersect  cd,  and  let  the  point 
of  intersection  be  taken  as  the  point  at  which  CD  is  resolved 
into  CO  and  OD ;  the  lines  of  action  of  these  forces  are  then 
co  and  od.  In  like  manner  choose  do,  oe,  intersecting  on  de,  as 
the  lines  of  action  of  DO,  OE.  If  this  is  done,  the  forces  OB, 
BO  will  neutralize  each  other  and  may  be  omitted  from  the 
system;  also  the  pairs  OC,  CO,  and  OD,  DO.  Hence,  there 
remain  only  the  two  forces  represented  in  magnitude  and 
direction  by  AO,  OE,  and  in  lines  of  action  by  ao,  oe. 
Their  resultant  is  given  in  magnitude  and  direction  by  AE, 
and  its  line  of  action  is  ae,  drawn  parallel  to  AE  through  the 


COMPOSITION   OF   NON-CONCURRENT   FO 

point  of  intersection  of  oa  and  oe ;  and  this  is  alsVtl 
ant  of  the  given  system. 

By  carefully  following  through  this  construction  the  student 
will  be  able  to  reduce  it  to  a  mechanical  method,  which  can 
be  readily  applied  to  any  system. 

28.  Funicular  Polygon.  —  The  polygon  whose  sides  are  oa,  ob, 
oc,  od,  oe,  is  called  a.  funicular  polygon*  for  the  given  forces. 

Since  the  point  at  which  the  two  components  of  AB  are 
assumed  to  act  may  be  taken  anywhere  on  the  line  ab,  there 
may  be  any  number  of  funicular  polygons  with  sides  parallel 
to  oa,  ob,  etc.  Again,  if  O  is  taken  at  a  different  point,  there 
may  be  drawn  a  new  funicular  polygon  starting  at  any  point 
of  ab ;  and  by  changing  the  starting  point  any  number  of 
funicular  polygons  may  be  drawn  with  sides  parallel  to  the 
new  directions  of  OA,  OB,  etc.  Moreover,  different  force 
and  funicular  polygons  may  be  obtained  by  changing  the  order 
in  which  the  forces  are  taken. 

It  may  be  proved  geometrically  that  for  every  possible  funic- 
ular polygon  drawn  for  the  same  system  of  forces,  the  last 
vertex,  determined  by  the  above  method,  will  lie  on  the  same 
line  parallel  to  the  closing  side  of  the  force  polygon  (as  ae, 
Fig.  9).  Such  a  proof  is  outside  the  scope  of  this  work. 
The  truth  of  the  proposition  may,  however,  be  shown  from 
the  principles  of  statics.  For  if  it  were  not  true,  it  would 
be  possible  by  the  above  method  to  find  two  or  more  forces, 
having  different  lines  of  action,  which  are  equivalent  to  each 
other,  because  each  is  equivalent  to  the  given  system.  But 
this  is  impossible. 

29.  Examples.  —  i.   Choose  five  forces,  assigning  the  magni- 
tude,   direction,    and   line    of   action    of   each,    and    find    their 
resultant   by  constructing   the   force   and   funicular    polygons. 

2.  Draw  a  second  funicular  polygon,  using  the  same  point 
O  in  the  force  diagram. 

*Also  called  equilibrium  polygon. 


jg  GRAPHIC   STATICS. 

3.  Draw  a  third  funicular  polygon,  choosing  a  new  point  O. 

4.  Solve  the  same  problem,  taking  the  forces  in  a  different 
order. 

30.  Definitions.  —  The  point  O  (Fig.  9)  is  called  the  pole  of 
the  force   polygon.     The   lines   drawn   from   the   pole  to   the 
vertices  of  the  force  polygon  may  be  called  rays.     The  sides 
of  the  funicular  polygon  are  sometimes  called  strings. 

Each  ray  in  the  force  diagram  is  parallel  to  a  corresponding 
string  in  the  space  diagram.  As  a  mechanical  rule,  it  should 
be  remembered  that  the  two  rays  drawn  to  the  extremities  of  the 
line  representing  any  force  are  respectively  parallel  to  the  two 
strings  which  intersect  on  the  line  of  action  of  that  force. 

The  rays  terminating  at  the  extremities  of  any  side  of  the 
force  polygon  represent  in  magnitude  and  direction  two  com- 
ponents that  may  replace  the  force  represented  by  that  side ; 
while  the  corresponding  strings  represent  the  lines  of  action  of 
these  components.  Thus  (Fig.  9)  the  force  represented  by  BC 
may  be  replaced  by  two  forces  represented  by  BO,  OC,  acting 
in  the  lines  bo,  oc. 

The  pole  distance  of  any  force  is  the  perpendicular  distance 
from  the  pole  to  the  line  representing  that  force  in  the  force 
diagram.  It  may  evidently  be  considered  as  representing  the 
resolved  part,  perpendicular  to  the  given  force,  of  either  of  the 
components  represented  by  the  corresponding  rays.  Thus,  OM 
(Fig.  9)  is  the  pole  distance  of  AB ;  and  OM  represents  the 
resolved  part,  perpendicular  to  AB,  of  either  OA  or  OB.  The 
student  should  notice  particularly  that  the  pole  distance  repre- 
sents a  force  magnitude  and  not  a  length. 

31.  Forces  not  Possessing  a  Single  Resultant.  —  It  may  hap- 
pen that  the  first  and  last  sides  of  the  funicular  polygon  are 
parallel,  so  that  the  above  construction  fails  to-  give  the  line  of 
action  of  the  resultant.     This  will  be  the  case  if  the  pole  is 
chosen  on  the  line  AE  (Fig.  9),  because  the  first  and  last  sides 
of  the  funicular   polygon  are  respectively  parallel  to  OA  and 


COMPOSITION    OF   NON-CONCURRENT 

OE.  The  difficulty  will,  in  this  case,  be  avoidefl  by  tstilirtgF  l&e 
pole  at  some  point  not  on  the  line  AE.  But  irr^ne  particular 
case  OA  and  OE  will  be  parallel,  wherever  tlie  pute+sire'  taken. 
By  inspection  of  the  force  diagram  it  is  seen  that  this  can 
occur  only  when  the  point  E  coincides  with  A.  In  this  case 
AO  and  OE  represent  equal  and  opposite  forces;  and  unless, 
their  lines  of  action  coincide,  they  cannot  be  combined  into  a 
simpler  system.  If  their  lines  of  action  are  coincident,  the 
forces  neutralize  each  other  and  the  resultant  is  zero  ;  if  not, 
the  system  reduces  to  a  couple  (Art.  7).  If  the  lines  of  action 
of  AO  and  OE  coincide,  they  may  still  be  regarded  as  forming 
a  couple,  its  arm  being  zero.  Therefore, 

If  the  force  polygon  for  any  system  of  forces  closes,  the  result- 
ant is  a  couple. 

Now,  it  is  evident  that  by  shifting  the  starting  point  for  the 
funicular  polygon,  the  lines  of  action  of  AO  and  OE  will  be 
shifted ;  and  by  taking  a  new  pole,  their  magnitude,  or  direc- 
tion, or  both,  may  be  changed.  Hence,  there  may  be  found 
any  number  of  couples,  each  equivalent  to  the  same  system 
of  forces,  and  therefore  equivalent  to  each  other.  The  relation 
between  equivalent  couples  is  discussed  in  Art.  52. 

Example. — Assume  five  forces  whose  force  polygon  closes, 
the  lines  of  action  being  taken  at  random.  Draw  two  funicular 
polygons,  using  the  same  pole,  and  a  third,  using  a  different 
pole ;  thus  reducing  the  given  system  to  an  equivalent  couple 
in  three  different  ways. 

32.  Resultant  Force  and  Resultant  Couple.  —  From  the  above 
discussion  (Arts.  27  and  31),  it  is  evident  that  any  system  of 
complanar  forces,  acting  on  the  same  rigid  body,  is  equivalent 
either  to  a  single  force  or  to  a  couple.     In  other  words,  every 
system  of  complanar  forces  possesses  either  a  resultant  force  or  a 
resultant  couple.     (See  Art.  9.) 

33.  Comparison  of  Methods.  —  Of  the  methods  given  in  Arts. 
26  and  27,  for  finding  the  resultant  of  a  system  of  non-concur- 


rg  GRAPHIC    STATICS. 

rent  forces^the  first  is  a  special  case  of  the  second.  For  if  the 
pole  ir^Fig.  9  be  chosen  at  the  point  A,  and  the  first  string  be 
made  to  pass  through  the  point  of  intersection  of  ab  and  be,  the 
construction  becomes  identical  with  that  of  Fig.  8.  If  the  first 
method  is  employed,  we  are  liable  to  meet  the  difficulty  men- 
tioned in  Arts.  25  and  26,  that  some  of  the  required  points  of 
intersection  do  not  fall  within  convenient  limits. 

In  the  second  method,  since  the  pole  may  be  chosen  at  pleas- 
ure, the  rays  may  generally  be  caused  to  make  convenient  angles 
with  the  given  forces,  and  all  required  points  of  intersection  to 
fall  within  convenient  limits. 

34.  Closing  of  the  Funicular  Polygon.  —  Let  the  force  poly- 
gon be  drawn  for  any  given  forces,  and  let  A  and  E  be  the 
initial  and  final  points.     Suppose  a  funicular  polygon  drawn, 
corresponding  to  any  pole  O.     The  given  system  is  equivalent 
to  two  forces  represented  in  magnitude  and  direction  by  AO, 
OE,  their  lines  of  action  being  the  first  and  last  sides  of  the 
funicular  polygon.     In  general,  these  lines  of  action  will  not  be 
parallel;  but,  as  explained  in  Art.  31,  it  may  happen  that  they 
are  parallel.     And  if  parallel,  it  may  happen  that  they  coincide. 
In  this  case,  the  funicular  polygon  is  said  to  be  closed. 

§  2.     Equilibrium  of  Non-concurrent  Forces. 

35.  Conditions  of  Equilibrium.  —  From  the  foregoing  princi- 
ples the  conditions  of  equilibrium  for  any  system  of  complanar 
forces  may  be  deduced. 

Whatever  the  system  of  forces  may  be,  let  a  force  polygon  be 
drawn,  and  let  A  and  E  be  its  initial  and  final  points.  Choose 
a  pole  O,  and  draw  a  funicular  polygon.  The  system  is  thus 
reduced  to  two  forces  represented  in  magnitude  and  direction 
by  AO  and  OE;  their  lines  of  action  being  the  first  and  last 
sides  of  the  funicular  polygon.  In  order  that  there  may  be 
equilibrium,  these  two  forces  must  be  equal  and  opposite  and 
have  the  same  line  of  action.  In  order  that  they  may  be  equal 


EQUILIBRIUM   OF   NON-CONCURRENT 

and  opposite,  A  and  E  must  coincide.  In  order  ttfkt 
have  the  same  line  of  action,  the  first  and  last  sides  \f  the  funic- 
ular polygon  must  coincide.  That  is,  in  order  that  tner6  may 
be  equilibrium,  two  conditions  must  be  satisfied : 

(a)  The  force  polygon  must  close. 

(b}  The  funicular  polygon  must  close. 

Conversely,  if  the  force  polygon  closes  and  one  funicular  poly- 
gon also  closes,  the  system  is  in  equilibrium.  For,  if  the  force 
polygon  closes,  the  two  forces  AO  and  OE  are  equal  and  oppo- 
site; and  if  a  funicular  polygon  closes,  these  two  forces  have 
the  same  line  of  action  and  therefore  balance  each  other. 

It  follows  that  if  the  force  polygon  and  one  funicular  polygon 
are  closed,  all  funicular  polygons  will  close. 

36.  Auxiliary  Conditions  of  Equilibrium.  —  If  any  system  of 
forces  in  equilibrium  be  divided  into  two  groups,  the  resultants 
of  the  two  groups  are  equal  and  opposite  and  have  the  same 
line  of  action. 

Particular  case.  —  If  three  forces  are  in  equilibrium,  their 
lines  of  action  must  meet  in  a  point,  or  be  parallel.  For,  if 
the  lines  of  action  of  two  of  the  forces  intersect,  their  result- 
ant must  act  in  a  line  passing  through  the  point  of  inter- 
section. But  this  resultant  must  be  equal  and  opposite  to 
the  third  force  and  have  the  same  line  of  action. 

37.  General  Method  of  Solving  Problems  in  Equilibrium.  — 

The  principles  of  Art.  35  furnish  a  general  method  of  solv- 
ing, graphically,  problems  relating  to  the  equilibrium  of  corn- 
planar  forces  acting  on  a  rigid  body.  The  problems  to  be 
solved  will  always  be  of  the  following  kind  : 

A  body  is  in  equilibrium  under  the  action  of  forces,  some 
of  which  are  completely  known,  and  others  wholly  or  partly 
unknown.  It  is  required  to  determine  the  unknown  elements. 
The  required  elements  may  be  either  the  magnitudes  or  the 
lines  of  action  of  the  forces. 


20 


GRAPHIC   STATICS. 


The  general  method  of  procedure  is  as  follows :  First,  draw 
the  force  polygon  and  funicular  polygon  so  far  as  possible 
from  the  given  data ;  then  complete  them,  subject  to  the 
condition  that  both  polygons  must  close. 

This  general  method  will  be  illustrated  by  the  solution  of 
several  problems  of  frequent  occurrence.  We  may  here  meet 
with  both  determinate  and  indeterminate  problems  (Art.  20). 
In  order  that  a  problem  may  be  determinate,  the  given  data, 
together  with  the  condition  that  the  force  polygon  and  funic- 
ular polygon  are  to  close,  must  be  just  sufficient  to  fully 
determine  these  figures.  There  are  many  possible  cases  fur- 
nishing determinate  problems.  Some  of  these  cannot  readily 
be  solved  graphically.  In  the  following  articles  are  treated 
three  important  cases  to  which  the  general  method  above 
outlined  is  well  adapted. 

38.  Problems  in  Equilibrium.  I. — A  rigid  body  is  in  equi- 
librium'under  the  action  of  a  system  of  parallel  forces,  all 
known  except  two,  these  being  unknown  in  magnitude  and 
direction,  but  having  known  lines  of  action.  It  is  required 
to  fully  determine  the  unknown  forces. 

Let  the  known  lines  of  action  of  five  forces  in  equilibrium 
be  abt  be,  cd,  de,  ea  (Fig.  10),  and  let 

I  I  AB,  BC,  CD  represent  the  known 

a  6  b  \c  a  e  c  d  de  forces  in  magnitude  and  direction, 

|*^ r — _          _  while  DE  and  EA  are  at  first  un- 

known. The  force  polygon,  so  far 
as  it  can  be  drawn  from  the  given 
data,  is  the  straight  line  ABCD. 
Choose  a  pole  O  and  draw  rays  OA, 
OB,  OC,  OD.  Parallel  to  these  draw 
in  the  space  diagram  the  strings  oa, 
obt  oc,  od,  four  successive  sides  of  the 
funicular  polygon.  This  much  can 
be  drawn  from  the  data  given.  We  must  now  complete  the 


EQUILIBRIUM   OF   NON-CONCURRENT   FORCES.  2I 


two  polygons  and  cause  both  to  close.  In  the  funicular  poly- 
gon, but  one  side  remains  to  be  drawn ;  and  in  the  force 
polygon  but  one  vertex  remains  unknown.  If  the  forces  are 
taken  in  the  order  AB,  BC,  CD,  DE,  EA,  the  unknown  vertex 
in  the  force  polygon  is  the  one  to  be  marked  E ;  the  unknown 
side  of  the  funicular  polygon  is  oe,  and  is  to  be  parallel  to 
the  line  OE.  But  the  string  oe  must  pass  through  the  inter- 
section of  od  with  de,  and  also  through  the  intersection  of 
oa  with  ae.  Hence,  since  these  intersections  are  both  known, 
oe  can  be  drawn  at  once  as  shown  in  the  figure;  and  then 
OE  can  be  drawn  parallel  to  oe.  This  fixes  the  point  E ; 
and  DE  and  EA  represent,  in  magnitude  and  direction,  the 
required  forces  whose  lines  of  action  are  de,  ea. 

39.  Problems  in  Equilibrium.  II.  —  A  rigid  body  is  in 
equilibrium  under  the  action  of  a  system  of  non-parallel  forces, 
all  known,  except  two;  of  these,  the  line  of  action  of  one 
and  the  point  of  application  of  the  other  are  given.  It  is 
required  to  completely  determine  the  unknown  forces. 

Let  the  system  consist  of  five  forces  to  be  represented  in 
the  usual  way ;  in  magnitude  and  direction  by  lines  marked 
AS,  BC,  CD,  DE,  EA  ;  and  in  lines  of  action  by  ab,  be,  cd, 
de,  ea.  Of  these  lines  let  be,  cd,  de,  ea  (Fig.  u),  be  given,  and 
let  L  be  a  given  point  of  ab.  Also  let  BC,  CD,  and  DE  be 
known,  while  EA,  AB  are  unknown.  First,  draw  the  force 
polygon  as  far  as  possible,  giving  B,  C,  D,  E,  four  consecutive 
vertices  of  the  force 
polygon.  The  side  EA 
must  be  parallel  to  ea, 
but  its  length  is  un- 
known, hence  the  ver- 
tex A  of  the  force 
polygon  cannot  be 
fixed.  Next,  draw  the 
funicular  polygon  so 


Fig.  11 


22  GRAPHIC    STATICS. 

far  as  possible  from  the  giver,  data.  Choose  the  pole  O,  and 
draw  the  rays  OB,  OC,  OD,  OE.  The  remaining  ray,  OA, 
cannot  yet  be  drawn.  Now,  in  the  funicular  polygon,  the 
sides  o by  oc  must  intersect  on  be ;  oc  and  o d  must  intersect 
on  cd\  od 'and  oe  must  intersect  on  de;  oe  and  oa  must  inter- 
sect on  ea  ;  and  oa  and  ob  must  intersect  on  ab.  Since  L 
is  the  only  known  point  of  the  line  ab,  let  this  point  be 
taken  as  the  point  of  intersection  of  oa  and  ob.  Now  draw 
ob  through  L  parallel  to  OB,  and  successively  oc,  od,  oc,  par- 
allel to  OC,  OD,  OE.  We  may  now  draw  oa,  joining  L  with 
the  point  in  which  oe  intersects  ea.  This  completes  the  funic- 
ular polygon.  The  force  polygon  can  now  be  completed  as 
follows  :  From  O  draw  a  line  parallel  to  oa  and  from  E  a  line 
in  the  known  direction  of  EA  ;  their  intersection  determines 
A.  This  determines  both  EA  and  AB,  and  the  force  polygon 
is  completely  known.  The  line  of  action  ab  may  now  be 
drawn  through  L  parallel  to  AB,  and  the  forces  are  completely 
determined. 

40.  Problems  in  Equilibrium.  III. — A  rigid  body  is  in 
equilibrium  under  the  action  of  any  number  of  forces,  of 
which  three  are  known  only  in  line  of  action  ;  the  remaining 
forces  being  completely  known.  It  is  required  to  determine 
the  unknown  forces. 

Let  the  given  forces  be  six  in  number,  their  lines  of  action 
being  represented  by  ab,  be,  cd,  de,  ef,  fa  (Fig.  12),  and  let  AB, 
BC,  and  CD  be  known,  while  the  remaining  three  forces  are  un- 
known in  magnitude  and  direction.  The  known  data  make  it 
possible  to  draw  at  once  three  sides  of  the  force  polygon, 
namely,  AB,  BC,  CD;  and  the  four  rays  OA,  OB,  OC,  OD, 
from  any  assumed  pole  O.  Also,  four  sides  of  the  funicular 
polygon,  oa,  ob,  oc,  od,  may  be  drawn  at  once.  But  oe  and  of 
are  unknown ;  also  DE,  EF,  and  FA  in  the  force  polygon. 

Since  any  side  of  the  funicular  polygon  can  be  drawn  through 
any  chosen  point,  let  the  polygon  be  started  by  drawing  oa 


EQUILIBRIUM   OF   NON-CONCURRENT   FORC 

through  the  intersection  of  efand  fa.  We  may  thenXftraw  suc- 
cessively ob,  oc,  od.  Now,  of  is  unknown  in  dire< 
is  to  be  drawn  through  the  intersection  of  oa  and  fa,  hence, 
whatever  its  direction,  it  intersects  ef  in  the  same  point  (since 
oa  was  drawn  through  the  intersection  of  ef  and  fa).  Hence, 
the  two  vertices  of  the  funicular  polygon  falling  on  ef  and  fa 
coincide  at  the  intersection  of  these  two  lines.  We  may  there- 


fore  draw  oe  through  this  point  and  also  through  the  point 
already  found  by  the  intersection  of  od  and  de.  Now  draw  a 
line  from  O  parallel  to  oe,  and  from  D  a  line  parallel  to  de ; 
their  intersection  is  E.  Draw  from  E  a  line  parallel  to  ef,  and 
from  A  a  line  parallel  to  fa ;  their  intersection  determines  F. 
The  force  polygon  is  now  completely  drawn,  and  DE,  EF,  FA 
represent,  in  magnitude  and  direction,  the  required  forces.  The 
remaining  ray  OF  and  the  corresponding  string  of  may  now  be 
drawn,  but  are  not  needed. 

The  construction  might  have  been  made  equally  well  by 
choosing  the  intersection  of  de  and  ef  as  the  starting  point, 
since  two  vertices  of  the  funicular  polygon  may  be  made  to 
coincide  at  that  point.  Or,  the  point  of  intersection  of  de  and 
of  might  be  chosen  ;  but  in  that  case,  the  order  of  the  forces 
should  be  so  changed  as  to  make  de  and  af  consecutive.  If 
this  were  done  the  figures  should  be  relettered  to  agree  with 


24  GRAPHIC    STATICS. 

the  order  in  which  the  forces  were  taken.  It  may  be  noticed 
that  the  direction  of  the  string  first  drawn  may  be  chosen  arbi- 
trarily and  the  pole  so  taken  as  to  correspond  to  the  direction 
chosen.  This  is  important  in  the  treatment  of  the  following 
special  case. 

Case  of  inaccessible  points  of  intersection.  —  It  may  happen 
that  the  lines  of  action  de,  ef,  and  fa  have  no  point  of  intersec- 
tion within  convenient  limits.  When  this  is  the  case,  the 
method  just  given  may  still  be  applied,  but  involves  the  geo- 
metrical problem  of  drawing  a  line  through  an  inaccessible  point. 
For  example,  if  ef  and  fa  intersect  beyond  the  limits  of  the 
drawing,  as  shown  in  Fig.  13,  we  may  proceed  as  follows  : 
Choose  some  point  of  ab  and  from  it  draw  a  line  through  the 
point  of  intersection  of  ef  and  fa.  (This  can  be  done  by  a 
method  to  be  explained  presently.)  Let  this  line  be  oa.  From 
the  known  point  A  of  the  force  polygon  draw  a  line  parallel  to 


Fig.  13 

oa  and  choose  a  pole  upon  it.  Draw  the  rays  OB,  OC,  OD; 
then  the  corresponding  strings  in  the  order  ob,  oc,  od.  From 
the  point  in  which  od  intersects  de  draw  a  line  to  the  inaccessible 
point  of  intersection  of  ef  and  fa ;  this  will  be  oe.  The  force 
polygon  can  now  be  completed  just  as  in  the  preceding  case. 

A  line  may  be  drawn  through  the  inaccessible  point  of  inter- 
section of  two  given  lines  by  the  following  method  :  Let  AC, 


EQUILIBRIUM   OF   NON-CONCURRENT   FORCES.  25 

BD  (Fig.  14),  be  the  given  lines,  and  let  it  be  required  to  draw 

a  line  through  their  point  of  intersection  from  some  point  P. 

Draw  PA  intersecting  AC  and 

BD,  and  from  C,  any  point  of 

AC,  draw  a  line  CQ  parallel  to 

PA.     From  E,  a  point  of  BD, 

draw  EA,  EP  ;    also  draw   CF 

parallel  to  AE,  and  FQ  parallel 

to  EP.      Then  PQ  is  the  line 

required.     For,   by   the    similar 

triangles,  it  is  easily  shown  that  Fis- 

A  R      ("C* 

—  = ;  which  proves  that  AC,  BG,  and  PQ  meet  in  a  point. 

Exception. — If  the  lines  of  action  of  the  three  unknown 
forces  meet  in  a  point  (or  are  parallel),  the  problem  is  impos- 
sible of  solution,  unless  it  happens  that  the  resultant  of  the 
known  forces  acts  in  a  line  through  the  point  of  intersection 
of  those  three  lines  of  action  (or  parallel  to  them) ;  in  which 
case  the  problem  is  indeterminate.  These  cases  will  not  be 
discussed  here. 

Remark.  —  In  the  problems  treated  in  this  and  the  two  pre- 
ceding articles,  it  will  be  noticed  that  the  forces  should  be  taken 
in  such  an  order  that  those  which  are  completely  known  are 
consecutive.  Otherwise  the  number  of  unknown  lines  in  the 
force  and  funicular  polygons  will  be  increased.  [For  an- 
other method  of  solving  this  problem,  see  Levy's  "  Statique 
Graphique."] 

41.  Examples.  —  i.  A  rigid  beam  AB  rests  horizontally 
upon  supports  at  A  and  B,  and  sustains  loads  as  follows: 
Its  own  weight  of  100  Ibs.  acting  at  its  middle  point  ;  a  load 
of  50  Ibs.  at  C ;  a  load  of  60  Ibs.  at  D ;  and  a  load  of  80  Ibs. 
at  E.  The  successive  distances  between  A,  C,  D,  E,  and  B 
are  4  ft.,  6  ft.,  7  ft.,  and  10  ft.  Required  the  upward  pressures 
on  the  beam  at  the  supports. 


26  GRAPHIC   STATICS. 

2.  A  rigid  beam  AB  is  hinged  at  A,  and  rests  horizontally 
with  the  end  B  upon  a  smooth  horizontal  surface.     The  beam 
sustains  loads  as  in  Example  I,  and  an  additional  force  of  40 
Ibs.  is  applied  at  the  middle  point  at  an  angle  of  45°  with  the 
bar  in  an  upward  direction.     Required  the  pressures  upon  the 
beam  at  A  and  B.     [The  pressure  at  A  may  have  any  direc- 
tion, while  the  pressure  at  B  must  be  vertically  upward,  i.e., 
at  right  angles  to  the  supporting  surface.     Hence,  this  is  a 
particular  case  of  Problem  II.] 

3.  Let  the  end  B  of  the  bar  rest  against  a  smooth  surface 
making   an   angle  of  30°  with   the   horizontal ;  the   remaining 
data  being  as  in  Example  2. 

4.  A  rigid  bar  2  ft.  long  weighs  10  Ibs.,  its  center  of  gravity 
being  8  inches  from  one  end.     The  bar  rests  inside  a  smooth 
hemispherical  bowl   of    15  inches  radius.     What  weight  must 
be  applied  at  the  middle  point  in  order  that  the  bar  may  rest 
when   making   an   angle   of    15°   with    the   horizontal?     Also, 
what  are  the  reactions  at  the  ends  ? 

5.  A  uniform  bar  20  inches   long,   weighing   10  Ibs.,    rests 
with   one  end   against  a  smooth   vertical  wall  and  the  other 
end  overhanging  a  smooth  peg   10  inches  from  the  wall.     A 
weight  P  is  suspended  from  the  end  so  that   the   bar   is   in 
equilibrium  when  making  an  angle  of  30°  with  the  horizontal. 
Find  P,  and  the  pressures  exerted  on  the   bar   by   the   wall 
and  peg. 

42.  Special  Methods.  —  Certain  problems  can  be  treated 
more  simply  by  other  methods  than  by  the  general  method 
of  constructing  the  force  and  funicular  polygons.  This  is 
sometimes  true  of  the  following : 

Problem.  —  A  rigid  body  is  held  in  equilibrium  by  four 
forces  acting  in  known  lines,  only  one  being  known  in  magni- 
tude and  direction.  It  is  required  to  completely  determine 
the  three  remaining  forces.  (See  Fig.  15.) 

Let  the  four  forces  have  lines  of  action  ab,  be,  cd,  da,  and  let 


EQUILIBRIUM   OF   NON-CONCURRENT   FORCES.  2/ 

AB  be  drawn  representing  the  magnitude  and  direction  of  the 
known  force.  Now  the  resultant  of  the  forces  whose  lines 
of  action  are  da  and  ab  must  act  in  a  line  passing  through 
M,  the  point  of  intersection  of  these  lines  ;  and  the  resultant 
of  the  other  two  forces  must  act  in  a  line  passing  through 
N,  the  point  of  intersection  of  be  and  cd.  But  the  two 


Fig.  15 


resultants  must  be  equal  and  opposite  and  have  the  same 
line  of  action,  else  there  could  not  be  equilibrium.  (Art.  36.) 
Hence,  each  must  act  in  the  line  MN.  Draw  BD  parallel 
to  MN,  and  AD  parallel  to  ad;  the  point  D  being  determined 
by  their  intersection.  Then  DA  represents  in  magnitude  and 
direction  the  force  acting  in  da,  and  DB  the  resultant  of 
DA  and  AB.  But  BD  must  represent  the  resultant  of  the 
two  remaining  forces ;  hence  these  two  forces  are  represented 
by  BC  and  CD  drawn  from  B  and  D  parallel  respectively  to 
be  and  cd. 

This  problem  is  a  special  case  of  that  treated  in  Art.  40. 
But  the  construction  here  given  will  in  many  cases  be  the 
simpler  one. 

Example.  —  A  rigid  body  has  the  form  of  a  square  ABCD, 
the  side  AB  being  horizontal,  and  BC  vertical.  The  weight 
of  the  body  is  100  Ibs.,  its  center  of  gravity  being  at  the  inter- 
section of  the  diagonals.  It  is  held  in  equilibrium  by  three 
forces  as  follows  :  Pv  acting  at  C  in  the  line  AC;  PI  acting  at 
D  in  the  line  AD ;  and  Ps  applied  at  B  and  acting  in  the  line 
joining  B  with  the  middle  point  of  AD.  Required  to  com- 
pletely determine  P1;  P2,  and  Ps. 


28 


GRAPHIC   STATICS. 


§  3.'  Resolution  into  Non-concurrent  Systems. 


43.  To  Replace   a   Force  by  Two  Non-concurrent  Forces. — 
This  may  be  done  in  an  infinite  number  of  ways.     The  lines 
of  action  of  the  two  components  must  intersect  at  a  point  on 
the  line  of  action  of  the  given  force,  and  they  must  further 
satisfy  the  same  conditions  as  concurrent  forces.     (Art.  21.) 

44.  To  Replace  a  Force  by  More  Than  Two  Non-concurrent 
Forces. — This  maybe  done  by  first  resolving  the  given  force 
into  two  forces  by  the  preceding  article,  and  then  resolving 
one  or  both  of  its  components  in  the  same  way.     This  problem 
and  that  of  Art.  43   are  indeterminate.     (See  note,  Art.   20.) 
To   make   such   a   problem    determinate,   something   must   be 
specified  regarding  the  magnitudes  and  lines  of  action  of  the 
required  components.     We  shall  consider  some  of  the  particular 
cases  which  are  of  frequent  use  in  the  treatment  of  practical 
problems. 

45.  Resolution  of  a  Force  into  Two  Parallel  Components.  — 
Let  it   be  required  to  resolve  a  force    into   two    components 
having  given  lines  of  action  parallel  to  its  own. 

If  the  given  force  be  reversed  in  direction,  it  will  form  with 
the  required  components  a  system  in  equilibrium.  The  com- 
ponents may  then  be  determined  by  the  method  of  Art.  38. 

Example.  —  Let  the  student  solve  two  particular  cases  of 
this  problem,  taking  the  line  of  action  of  the  given  force  (i) 
between  those  of  the  components,  (2)  outside  those  of  the 
components. 

46.  Resolution  of  a  Force  into  Three  Components.  —  Problem. 
—  To  resolve  a  force  into  three  components  having  known  lines 
of  action. 

If  the  given  force  be  reversed  in  direction,  it  will  form,  with 
the  required  forces,  a  system  in  equilibrium.  Hence  these 
forces  may  be  determined  by  either  of  the  two  methods  given 


RESOLUTION  INTO  NON-CONCURRENT  SYST 


in    Arts.    40   and   42.      Or,    the   following    reaso 
employed,  leading  to  the  same  construction  as  tha 

Let  AD  (Fig.  16)  represent  the  given  force 
and  direction,  and  ad  its  line  of  action  ;  the  lines  of  action  of 
the  components  being  given  as  ab,  be,  cd.  Since  the  given 
force  may  be  assumed  to  act  at  any  point  in  the  line  ad,  let  its 
point  of  application  be  taken  at  M,  the  point  of  intersection  of 


ad  and  cd.  Resolve  it  into  two  components  acting  in  the  lines 
cd  and  MN,  N  being  the  point  of  intersection  of  ab  and  be. 
These  components  are  represented  in  magnitude  and  direction 
by  AC,  CD.  Let  AC,  acting  in  the  line  MN  (also  marked  ac], 
be  resolved  into  components  having  lines  of  action  ab,  be. 
These  components  are  given  in  magnitude  and  direction  by 
AB,  BC,  drawn  parallel  respectively  to  ab,  be.  Hence,  the 
given  force,  acting  in  ad,  is  equivalent  to  the  three  forces 
represented  in  magnitude  and  direction  by  AB,  BC,  CD,  acting 
in  the  lines  ab,  be,  cd. 

If  the  line  of  action  of  the  given  force  does  not  intersect  any 
one  of  the  given  lines  ab,  be,  cd,  within  the  limits  of  the  draw- 
ing, it  may  be  replaced  by  two  components  ;  then  each  may  be 
resolved  in  accordance  with  the  above  method,  and  the  results 
combined.  If  the  three  lines  ab,  be,  cd,  are  all  parallel  to  ad, 
or  if  the  four  lines  intersect  in  a  point,  the  problem  is  inde- 
terminate. This  is  evident  from  the  preceding  articles,  since, 
by  methods  already  given,  a  part  of  AD  can  be  replaced  by 
two  components  acting  in  any  two  of  the  given  lines,  as  ab, 
be ;  and  the  remaining  part  by  two  components  acting  in  ab,  cd; 


30  GRAPHIC   STATICS. 

or  in  be,  cd.     This   construction  evidently  admits    of   infinite 
variation. 

For  another  method  of  solving  the  above  problem,  see  Clarke's 
"Graphic  Statics,"  p.  16. 

§  4.    Moments  of  Forces  and  of  Couples. 

47.  Moment  of  a  Force.  —  Definition.  —  The  moment  of  a 
force  with  respect  to  a  point  is  the  product  of  the  magnitude  of 
the  force  into  the  perpendicular  distance  of  its  line  of  action 
from  the  given  point.  The  moment  of  a  force  with  respect  to 
an  axis  perpendicular  to  the  force  is  the  product  of  the  magni- 
tude of  the  force  by  the  perpendicular  distance  from  the  axis 
to  the  line  of  action  of  the  force. 

If  the  moment  is  taken  with  respect  to  a  point,  that  point  is 
called  the  origin  of  moments.  The  perpendicular  distance  from 
the  origin,  or  axis,  to  the  line  of  action  of  the  force  is  called 
the  arm. 

Rotation  tendency  of  a  force.  —  The  moment  of  a  force 
measures  its  tendency  to  produce  rotation  about  the  origin,  or 
axis.  Thus,  if  a  rigid  body  is  fixed  at  a  point,  but  free  to  turn 
about  that  point  in  a  given  plane,  any  force  acting  upon  it  in 
that  plane  will  tend  to  cause  it  to  rotate  about  the  fixed  point. 
The  amount  of  this  tendency  will  be  proportional  both  to  the 
magnitude  of  the  force  and  to  the  distance  of  its  line  of  action 
from  the  given  point ;  that  is,  to  the  moment  of  the  force  with 
respect  to  the  point,  as  above  defined. 

Rotation  in  any  plane  may  have  either  of  two  opposite 
directions,  which  may  be  distinguished  from  each  other  by  signs 
plus  and  minus.  Rotation  with  the  hands  of  a  watch  supposed 
placed  face  upward  in  the  plane  of  the  paper  will  be  called 
negative,  and  the  opposite  kind  positive.  It  would  be  equally 
legitimate  to  adopt  the  opposite  convention,  but  the  method 
here  adopted  agrees  with  the  usage  of  the  majority  of  writers. 

The  sign  of  the  moment  of  a  force  is  regarded  as  the  same 
as  that  of  the  rotation  it  tends  to  produce  about  the  origin. 


MOMENTS   OF   FORCES   AND   OF   COUPLES.  31 

Moment  represented  by  the  area  of  a  triangle.  —  If  a  triangle 
be  constructed  having  for  its  vertex  the  origin  of  moments,  and 
for  its  base  a  length  in  the  line  of  action  of  a  force,  numerically 
equal  to  its  magnitude,  then  the  moment  of  the  force  is  numeri- 
cally equal  to  double  the  area  of  the  triangle.  This  follows  at 
once  from  the  definition  of  moment. 

48.  Moment  of  Resultant  of  Two  Non-parallel  Forces.  —  Propo- 
sition. —  The  moment  of  the  resultant  of  two  non-parallel  forces 
with  reference  to  a  point  in  their  plane  is  equal  to  the  algebraic 
sum  of  their  separate  moments  with  reference  to  the  same  point. 

In  Fig.  17,  let  AB,  BC,  and  AC  represent  in  magnitude  and 
direction  two  forces  and  their  resultant ;  and  let  ab,  be,  ac  be 
their  lines  of  action,  intersecting  in  a  .. 
point  N.  Choose  any  point  M  as  the  l  <v 
origin  of  moments.  Lay  off  NP  =  AB\ 
NQ  =  BC;  and  NR  =  AC.  Then  the 
moments  of  the  three  forces  are  re- 
spectively equal  to  double  the  areas  of 
the  triangles  MNP,  MNQ,  MNR. 
These  three  triangles  have  a  common 
side  MN,  which  may  be  considered  the 
base ;  hence  their  areas  are  propor- 
tional to  their  altitudes  measured  from  that  side.  These  alti- 
tudes are  PP',  QQ',  RR',  perpendicular  to  MN.  Now,  if  PP" 
is  parallel  to  MN,  P"R'  is  equal  to  PP' ;  also  RP"  equals 
QQ'  (since  they  are  homologous  sides  of  the  equal  triangles 
NQQ',  PRP"}.  Hence  RR'  =  PP'  +  QQ',  and  therefore 

Area  MNR  =  area  MNP + area  MNQ  ; 

which  proves  the  proposition. 

If  the  origin  of  moments  is  so  taken  that  the  moments  of 
AB  and  BC  have  opposite  signs,  the  demonstration  needs  modi- 
fication. The  student  should  attempt  the  proof  of  this  case  for 
himself. 


32  GRAPHIC    STATICS. 

49.  Moment  of  a  Couple.  —  Definition.  —  The  moment  of  a 
couple  about  any  point  in  its  plane  taken  as  an  origin  is  the 
algebraic  sum  of  the  moments  of  the  two  forces  composing  it 
with  reference  to  the  same  origin. 

Proposition.  —  The  moment  of  a  couple  is  the  same  for  every 
origin  in  its  plane,  and  is  numerically  equal  to  the  product  of 
the  magnitude  of  either  force  into  the  arm  of  the  couple. 
(Art.  7.) 

The  proof  of  this  proposition  can  readily  be  supplied  by  the 
student. 

50.  Moment  of  the  Resultant  of  Any  System.  —  Proposition. 
—  The  algebraic  sum  of  the  moments  of  any  given  complanar 
forces,  with  reference  to  any  origin  in  their  plane,  is  equal  to 
the  moment  of  their  resultant  force  or  resultant  couple  with 
reference  to  the  same  origin. 

The  construction  employed  in  proving  this  proposition  is 
similar  to  that  used  in  Art.  27,  which  the  student  may  profit- 
ably review  at  this  point.  Referring  to  Fig.  9,  let  the  given 
forces  be  represented  in  magnitude  and  direction  by  AB,  BC, 
CD,  DE,  and  in  lines  of  action  by  ab,  be,  cd,  de.  Let  the 
origin  of  moments  be  any  point  in  the  space  diagram.  As  in 
Art.  27,  replace  AB  by  AO,  OB,  acting  in  lines  ao,  ob;  replace 
BC  by  BO,  OC,  acting  in  lines  bo,  oc ;  replace  CD  by  CO,  OD, 
acting  in  lines  co,  od;  and  replace  DE  by  DO,  OE,  acting  in 
lines  do,  oe.  (For  brevity,  we  refer  to  a  force  as  AB,  meaning 
"the  force  represented  in  magnitude  and  direction  by  AB.") 

Now  by  Art.  48,  we  have,  whatever  the  origin, 

Moment  of  AB  =  moment  of  A O  +  moment  of  OB; 

"  BC=  "  "  BO+  "  "  OC; 
"  CD  =  "  "  CO+  "  "  OD; 
"  DE=  "  "  DO+  "  "  OE. 

Since  the  forces  represented  by  BO  and  OB  have  the  same 
line  of  action,  their  moments  are  numerically  equal  but  have 


MOMENTS  OF  FORCES  AND  OF  COUP 

opposite  signs ;  and  similar  statements  are  true 
OC,  and  of  DO  and  OD.  Hence,  the  addition  of  the  above 
four  equations  shows  that  the  sum  of  the  moments  of  AB, 
BC,  CD,  DE,  is  equal  to  the  sum  of  the  moments  of  AO 
and  OE.  Now  the  given  system  has  either  a  resultant  force 
or  a  resultant  couple.  In  the  first  case  the  resultant  of  the 
system  is  the  resultant  of  AO  and  OE,  and  its  moment  is 
equal  to  the  algebraic  sum  of  their  moments,  by  Art.  48. 
In  the  second  case  (which  occurs  only  when  E  coincides  with 
A),  the  resultant  couple  is  composed  of.  AO  and  OE  (which 
in  this  case  are  equal  and  opposite  forces),  and  the  moment 
of  the  couple  is,  by  definition,  equal  to  the  algebraic  sum  of 
the  moments  of  AO  and  OE.  Hence,  in  either  case,  the 
proposition  is  true. 

It  should  be  noticed  that  the  proof  here  given  applies  to 
systems  of  parallel  forces,  as  well  as  to  non-parallel  systems. 
The  proposition  of  Art.  48  could  be  extended  to  the  case  of 
any  number  of  forces,  by  considering  first  the  resultant  of 
two  forces,  then  combining  this  resultant  with  the  third  force, 
and  so  on  ;  but  the  method  would  fail  if,  in  the  process,  it 
became  necessary  to  combine  parallel  forces.  The  method 
here  adopted  is  not  subject  to  this  failure. 

51.  Condition  of  Equilibrium.  —  It  follows  from  what  has 
preceded,  that  if  a  given  system  is  in  equilibrium,  the  alge- 
braic sum  of  the  moments  must  be  zero,  whatever  the  origin. 
For,  in  case  of  equilibrium,  AO  and  OE  (Fig.  9)  are  equal  and 
opposite  and  have  the  same  line  of  action  ;  hence,  the  sum 
of  their  moments  (which  is  the  same  as  the  sum  of  the 
moments  of  the  given  forces)  is  equal  to  zero.  Conversely, 
If  the  algebraic  sum  of  the  moments  is  zero  for  every  origin, 
the  system  must  be  in  equilibrium.  For,  if  it  is  not,  there 
is  either  a  resultant  force  or  a  resultant  couple.  But  the 
moment  of  a  force  is  not  zero  for  any  origin  not  on  its  line 
of  action ;  and  the  moment  of  a  couple  is  not  zero  for  any 


34 


GRAPHIC    STATICS. 


origin..    For  a  fuller  discussion  of  the  conditions  of  equilibrium, 
see  Arts.  58  and  59. 

52.  Equivalent  Couples. — Proposition.  —  If  a  system  has  for 
its   resultant  a  couple,   it  is  equivalent  to  any  couple  whose 
moment  is   equal  to  the  sum  of  the  moments  of    the  forces 
of  the  system. 

For,  as  already  seen  (Art.  31),  when  the  resultant  is  a 
couple,  the  force  polygon  is  closed.  Let  the  initial  and  final 
points  of  the  force  polygon  coincide  at  some  point  A,  and 
let  O  be  the  pole.  Then  the  forces  of  the  resultant  couple 
are  represented  in  magnitude  and  direction  by  AO,  OA.  Since 
the  position  of  O  is  arbitrary,  the  force  AO  (or  OA)  may  be 
made  anything  whatever  in  magnitude  and  direction.  Also 
the  line  of  action  of  the  force  AO  may  be  taken  so  as  to  pass 
through  any  chosen  point.  Hence,  the  resultant  couple  may 
have  for  one  of  its  forces  any  force  whatever  in  the  plane  of 
the  given  system  ;  and  the  other  force  will  have  such  a  line 
of  action  that  the  moment  of  the  couple  will  be  equal  to  the 
sum  of  the  moments  of  the  given  forces. 

This  reasoning  is  equally  true  if  the  given  system  is  a 
couple.  Hence,  a  couple  is  equivalent  to  any  other  couple 
having  the  same  moment.  In  other  words,  all  couples  wJiose 
moments  are  equal  are  equivalent ;  and  conversely,  all  equiva- 
lent couples  have  equal  moments. 

[NOTE.  — The  construction  above  discussed  fails  if  all  forces  of  the  given  system 
are  parallel  to  the  direction  chosen  for  the  forces  of  the  resultant  couple.  For  then 
the  force  polygon  is  a  straight  line,  and  if  the  pole  is  chosen  in  that  line,  the 
strings  of  the  funicular  polygon  are  parallel  to  the  lines  of  action  of  the  forces,  and 
the  polygon  cannot  be  drawn.  But  in  this  case,  the  system  may  first  be  reduced  to  a 
couple  whose  forces  have  some  other  direction,  and  this  couple  may  be  reduced  to 
one  whose  forces  have  the  direction  first  chosen.  Hence,  the  proposition  stated 
holds  in  all  cases.] 

53.  Moment  of  a  System.  —  Definition.  — The  moment  of  a 
system  of  forces  is  the  algebraic  sum  of  the  moments  of  the 
forces  of  the  system. 


GRAPHIC   DETERMINATION   OF   MOMEN 

54.  Moments   of   Equivalent    Systems. — Propositu 
moments  of  any  two  equivalent  systems   of  complanar  forces 
with  respect  to  the  same  origin  are  equal. 

This  follows  immediately  from  the  preceding  articles.  For, 
if  the  two  systems  are  equivalent,  each  is  equivalent  to  the 
same  resultant  force  or  resultant  couple,  and  the  moments  of 
the  two  systems  are  therefore  each  equal  to  the  moment  of 
this  resultant  and  hence  to  each  other. 

§  5.    Graphic  Determination  of  Moments. 

55.  Proposition.  —  If,  through   any  point  in  the   space  dia- 
gram a  line  be  drawn  parallel  to  a  given  force,  the  distance 
intercepted  upon  it  by  the  two  strings  corresponding  to  that 
force,   multiplied  by  the  pole  distance  of   the  force,  is  equal 
to  the  moment  of  the  force  with  respect  to  the  given  point. 

By  the  strings  "  corresponding  to  "  a  given  force  are  meant 
the  two  strings  which  intersect  at  a  point  on  its  line  of  action. 

Let  AB  (Fig.  18)  represent  the  magnitude  and  direction  of 
a  force  whose  line  of  action  is  ab,  and  let  M  be  the  origin 
of  moments.     Let  O  be  the  pole, 
and  OK  (=//)  the  pole  distance 
of  the  given  force.     Draw  the 
strings  oa,   ob,  and    through  M 
draw  a  line  parallel  to  ab,  inter- 
secting oa  and  ob  in  P  and  Q.  Fig.  is 
Then  it  is  to  be  proved  that  the 
moment    of   the    given    force   with    respect   to  M  is  equal  to 

HY.PQ. 

Let  h  equal  the  perpendicular  distance  of  M  from  ab.  Then 
the  required  moment  is  AB  x  //.  But  since  the  similar  triangles 
OAB  and  RPQ  have  bases  AB,  PQ,  and  altitudes  H,  h,  respec- 
tively, it  follows  that 

J>Q  =  ~\  hence, 
which  proves  the  proposition. 


36  GRAPHIC   STATICS. 

It  should  be  noticed  that  PQ  represents  a  length,  while  H 
represents  a  force  magnitude.  Hence,  the  moment  of  the  given 
force  with  respect  to  M  is  equal  to  the  moment  of  a  force  H 
with  an  arm  PQ.  [It  may,  in  fact,  be  shown  that  the  given 
force  is  equivalent  to  an  equal  force  acting  in  the  line  PQ 
(whose  moment  about  M  is  therefore  zero),  and  a  couple  with 
forces  of  magnitude  H,  and  arm  PQ.  For  AB  acting  in  ab  is 
equivalent  to  AO  and  OB  acting  in  ao  and  ob  respectively. 
Also,  AO  acting  in  ao  is  equivalent  to  forces  represented  by 
AK  and  KO  acting  respectively  in  PQ  and  in  a  line  through 
P  at  right  angles  to  PQ  ;  and  OB  may  be  replaced  by  forces 
represented  by  OK  and  KB,  the  former  acting  in  a  line  through 
Q  perpendicular  to  PQ,  and  the  latter  in  PQ.  But  AK  and 
KB  are  equivalent  to  AB  ;  hence,  the  proposition  is  proved.] 

56.  Moment  of  the  Resultant  of  Several  Forces.  — The  mo- 
ment of  the  resultant  of  any  number  of  consecutive  forces  in 
the  force  and  funicular  polygons  may  be  found  by  a  method 
similar  to  that  just  described.  Thus,  let  Fig.  19  represent  the 

force  polygon  and  the 
funicular  polygon  for  six 
forces,  and  let  it  be  re- 
quired to  find  the  moment 
of  the  resultant  of  the  four 
forces  represented  in  the 
force  polygon  by  BC,  CD, 
DE,  and  EF,  with  respect 
to  any  point  M.  The  re- 
sultant of  the  four  forces 
is  represented  by  BF,  and 

acts  in  a  line  through  the  intersection  of  ob  and  of.  Through 
M  draw  a  line  parallel  to  BF,  intersecting  ob  and  of\r\P  and  Q 
respectively.  Then  PQ  multiplied  by  OK,  the  pole  distance  of 
BF,  gives  the  required  moment.  This  method  does  not  apply 
to  the  determination  of  the  moment  of  the  resultant  of  several 
forces  not  consecutive  in  the  force  polygon. 


SUMMARY   OF   CONDITIONS   OF   EQUILIBR 

57.  Moments  of  Parallel  Forces. — The  method  of 

and  56  is  especially  useful  when  it  is  desired  to  find  the  mo- 
ments of  any  or  all  of  a  system  of  parallel  forces  ;  since  with 
such  a  system  the  pole  distance  is  the  same  for  all  forces,  and 
the  moments  are  therefore  proportional  to  the  intercepts  found 
by  the  above  method. 

Example.  —  Assume  five  parallel  forces  at  random ;  choose 
an  origin,  and  determine  their  separate  moments,  also  the 
moment  of  their  resultant,  by  the  method  of  Arts.  55  and  56. 

§  6.    Summary  of  Conditions  of  Equilibrium. 

58.  Graphical  and  Analytical  Conditions  of  Equilibrium  Com- 
pared.—  It  has  been  shown  (Art.   35)  that   the  conditions  of 
equilibrium  for  a  system  of  complanar  forces  acting  on  a  rigid 
body  are  two  in  number  : 

(I)  The  force  polygon  must  close. 

(II)  Any  funicular  polygon  must  close. 

The  analytical  conditions  *  are  the  following  : 

(1)  The   algebraic  sum  of  the  resolved  parts  of  the  forces 
in  any  direction  must  be  zero. 

(2)  The  algebraic  sum  of  their  moments  for  any  origin  must 
be  zero.' 

The  condition  (i)  is  readily  seen  to  be  equivalent  to  (I). 
For  if  the  sides  of  the  force  polygon  be  orthographically 
projected  upon  any  line,  their  projections  will  represent  in 
magnitude  and  direction  the  resolved  parts  of  the  several 
forces,  parallel  to  the  line;  and,  if  the  force  polygon  is  closed, 
the  algebraic  sum  of  these  projections  is  zero,  whatever  the 
direction  of  the  assumed  line.  (See  Art.  22.)  It  may  also 
be  seen  that  condition  (II)  carries  with  it  (2).  For,  if  every 
funicular  polygon  closes,  the  system  is  equivalent  to  two  equal 
and  opposite  forces  having  the  same  line  of  action  (Arts.  31, 

*  See  Mine/tin's  Statics,  Vol.  I,  p.  114. 


38  GRAPHIC   STATICS. 

35,  and  50) ;  and  the  sum  of  the  moments  of  these  two  forces 
must  be  zero. 

A  further  comparison  may  be  made.  The  analytical  condi- 
tion (2)  carries  (i)  with  it  ;  and  similarly  the  graphical  condi- 
tion (II)  carries  with  it  the  condition  (I).  That  (2)  includes 
(i)  may  be  seen  as  follows  :  If  the  sum  of  the  moments  is  zero 
for  one  origin  Ml}  there  can  be  no  resultant  couple,  neither  can 
there  be  a  resultant  force  unless  with  a  line  of  action  passing 
through  Mv  If  the  sum  of  the  moments  is  zero  for  a 
second  origin  M2,  the  resultant  force,  if  one  exists,  must 
act  in  the  line  M^M*  If  the  sum  of  the  moments  is  zero 
also  for  a  third  origin  Mz,  not  on  the  line  MiM2,  there  can 
be  no  resultant  force.  It  follows  at  once  that  condition  (i) 
must  hold. 

That  (II)  includes  (I)  may  be  shown  as  follows  :  Let  A  and 
E  be  the  first  and  last  points  of  the  force  polygon,  and  choose 
a  pole  Oi.  Then,  if  the  funicular  polygon  closes,  OiA  and 
OiE  are  parallel.  Choose  a  second  pole  O2,  and,  if  the  funic- 
ular polygon  again  closes,  O2A  and  O2E  are  parallel.  AE 
must  then  be  parallel  to  OiO2,  unless  A  and  E  coincide. 
Now,  choose  a  third  pole  Os,  not  on  O\O%\  if  the  funicular 
polygon  for  this  pole  closes,  OZA  and  O3E  must  be  parallel. 
But  this  is  impossible  unless  A  and  E  coincide ;  that  is,  unless 
condition  (I)  holds. 

The  last  result  may  be  reached  in  another  way.  With  any 
pole  O  draw  a  funicular  polygon  and  suppose  it  to  close.  The 
system  is  thus  reduced  to  two  forces  acting  in  the  same  line 
oa.  Hence,  there  is  no  resultant  couple,  and  if  there  is  a 
resultant  force,  its  line  of  action  is  oa.  With  the  same  pole 
draw  a  second  funicular  polygon,  the  first  side  being  o'a', 
parallel  to  oa.  If  this  polygon  closes,  there  can  be  no  result- 
ant force,  for  if  one  existed  it  would  act  in  the  line  o'a' ;  and 
there  would  thus  be  two  resultant  forces,  acting  in  different 
lines  oa  and  o'a',  which  is  impossible. 


SUMMARY    OF   CONDITIONS    OF   EQUILIBRIUM.  39 

59.  Summary.  — It  is  now  evident  that  the  conditions  neces- 
sary to  insure  equilibrium  may  be  stated  in  several  different 
ways,  both  analytically  and  graphically.  To  summarize  : 

A.  Analytically  :  There  will  be  equilibrium  if  either  of  the 
following  conditions  is  satisfied  : 

(1)  The  sum  of  the  moments  is  zero  for  each  of  three  points 
not  in  the  same  line. 

(2)  The  sum  of  the  moments  is  zero  for  each  of  two  points, 
and  the  sum  of  the  resolved  parts  is  zero  in  a  direction  not 
perpendicular  to  the  line  joining  those  two  points. 

(3)  The  sum  of  the  moments  is  zero  for  one  point,  and  the 
sum  of  the  resolved  parts  is  zero  for  each  of  two  directions. 

B.  Graphically :  There  will  be  equilibrium  if  either  of  these 
three  conditions  is  satisfied  : 

(1)  A  funicular  polygon  closes  for  each  of  three  poles  not  in 
the  same  line. 

(2)  Two  funicular  polygons  close  for  the  same  pole. 

(3)  One  funicular  polygon  closes  and  the  force  polygon  closes. 


CHAPTER  III.     INTERNAL  FORCES  AND 
STRESSES. 

§  i .    External  and  Internal  Forces. 

60.  Definitions.  —  It  was  stated  in  Art.  3  that  every  force 
acting  upon  any  body  is  exerted  by  some  other  body.  In  what 
precedes,  we  have  been  concerned  only  with  the  effects  pro- 
duced by  forces  upon  the  bodies  to  which  they  are  applied. 
It  has  therefore  not  been  needful  to  consider  the  bodies  which 
exert  the  forces.  It  is  now  necessary  to  consider  forces  in 
another  aspect. 

The  forces  applied  to  any  particle  of  a  body  may  be  either 
external  or  internal. 

An  external  force  is  one  exerted  upon  the  body  in  question 
by  some  other  body. 

An  internal  force  is  one  exerted  upon  one  portion  of  the  body 
by  another  portion  of  the  same  body. 

It  is  important  to  note,  however,  that  the  same  force  may  be 
internal  from  one  point  of  view,  and  external  from  another. 
Thus,  if  a  given  body  be  conceived  as  made  up  of  two  parts, 
X  and  Y,  a  force  exerted  upon  X  by  Y  is  internal  as  regards 
the  whole  body,  but  external  as  regards  the  part  X.  Thus,  let 

AB  (Fig.  20)  represent 

-* f  f  & *Q  a  bar,  acted  upon  by  two 

A  B  forces  of  equal  magnitude 

JTig.  QO 

applied  at  the  ends  paral- 
lel to  the  length  of  the  bar,  in  such  a  way  as  to  tend  to  pull  it 
apart.  These  two  forces  are  exerted  upon  AB  by  some  other 

40 


EXTERNAL  AND   INTERNAL   FORCES. 

bodies   not   specified.      If   the   whole   bar  be   consic 

external   forces    acting    upon    it    are    simply   the    two    forces 

named. 

But  suppose  the  body  under  consideration  is  AC,  a  portion  of 
AB.  The  external  forces  acting  upon  this  body  are  (i)  a  force 
at  A,  already  mentioned,  and  (2)  a  force  at  C,  exerted  upon  AC 
by  CB.  This  latter  force  is  internal  to  the  bar  AB,  but  external 
to  AC. 

61.   Conditions  of  Equilibrium  Apply  to  External  Forces.  —  In 

applying  the  conditions  of  equilibrium  deduced  in  previous 
articles,  it  must  be  remembered  that  only  external  forces  are 
referred  to.  It  is  also  important  to  notice  that  the  principles 
apply  to  any  body  or  any  portion  of  a  body  in  equilibrium ;  and 
the  system  of  forces  in  every  case  must  include  all  forces  that 
are  external  to  the  body  or  portion  of  a  body  in  question. 

Thus,  if  the  bar  AB  (Fig.  20)  be  in  equilibrium  under  the 
action  of  two  opposite  forces  P  and  Q,  applied  at  A  and  B 
respectively,  as  shown  in  the  figure,  the  principles  of  equilibrium 
may  be  applied  either  to  the  whole  bar,  or  to  any  part  of  it,  as 
AC. 

(a)  For  the  equilibrium  of  the  whole  bar  AB,  we  must  have 
P=Q,  these  being  supposed  the  only  external  forces  acting  on 
the  bar. 

(b)  For  the  equilibrium  of  AC,  the  force  exerted  upon  AC  by 
CB  must  be   equal   and  opposite  to  P.     This  latter  force  is 
external  to  A  C,  though  internal  to  the  whole  bar. 

The  method  just  illustrated  is  of  frequent  use  in  the  investi- 
gation of  engineering  structures.  It  is  often  desired  to  deter- 
mine the  internal  forces  acting  in  the  members  of  a  structure, 
and  the  general  method  is  this :  Direct  the  attention  to  such  a 
portion  of  the  whole  structure  or  body  considered  that  the 
internal  forces  which  it  is  desired  to  determine  shall  be  external 
to  the  portion  in  question.  (See  Art.  67.) 


42 


GRAPHIC    STATICS. 


§  2.   External  and  Internal  Stresses. 


62.  Newton's  Third  Law.  —  Let  X  and  Y  be  any  two  portions 
of  matter  ;  then  if  X  acts  upon  Y  with  a  certain  force,  F  acts 
upon  X  with  a  force  of  equal  magnitude  in  the  opposite  direc- 
tion.    This  is  the  principle   stated  in  Newton's  third  law  of 
motion,  —  that  "  to  every  action  there  is  an  equal  and  contrary 
reaction."     It  is  justified  by  universal  experience. 

63.  Stress.  —  Definition.  —  Two  forces  exerted  by  two  por- 
tions of  matter  upon  each  other  in  such  a  way  as  to  constitute 
an  action  and  its  reaction,  make  up  a  stress. 

Illustrations.  —  The  earth  attracts  the  moon  with  a  certain 
force,  and  the  moon  attracts  the  earth  with  an  equal  and  opposite 
force.  The  two  forces  constitute  a  stress. 

Two  electrified  bodies  attract  (or  repel)  each  other  with  equal 
and  opposite  forces.  These  two  forces  constitute  a  stress. 

Any  two  bodies  in  contact  exert  upon  each  other  equal  and 
opposite  pressures  (forces),  constituting  a  stress. 

By  the  magnitude  of  a  stress  is  meant  the  magnitude  of  either 
of  its  forces. 

64.  External  and  Internal  Stresses.  —  It  has  been  seen  that 
two  portions  of  matter  are  concerned  in  every  stress.     Now  the 
two  portions  may  be  regarded  either  as  separate  bodies,  or  as 
parts  of  a  body  or  system  of  bodies  which  includes  both. 

A  stress  acting  between  two  parts  of  the  same  body  (or 
system  of  bodies)  is  an  internal  stress  as  regards  that  body 
or  system. 

A  stress  acting  between  two  distinct  bodies  is  an  external 
stress  as  regards  either  body. 

It  is  important  to  notice  that  the  same  stress  may  be  internal 
from  one  point  of  view,  and  external  from  another.  Thus,  if 
a  given  body  be  considered  as  made  up  of  two  parts  X  and  F, 
a  stress  exerted  between  X  and  F  is  internal  to  the  whole  body, 
but  external  to  either  X  or  F. 


EXTERNAL   AND   INTERNAL   STRESSES. 


43 


Illustration.  — Consider  a  body  AB  (Fig.  21)  resting  upon  a 
second  body  Y,  and  supporting  another  body  X,  as  shown.  If 
the  weight  of  the  body  AB  be  disregarded, 
the  forces  acting  upon  it  are  (i)  the  down- 
ward pressure  (say  P)  exerted  by  X  at  the 
surface  A,  and  the  upward  pressure  (say  Q) 
exerted  by  Fat  B  ;  these  forces  being  equal 
and  opposite,  since  the  body  is  in  equilibrium. 
Now  the  body  X  is  acted  upon  by  AB  with 
a  force  equal  and  opposite  to  P,  and  these 
two  forces  constitute  a  stress  which  is  external  to  AB.  There 
is  also  an  external  stress  exerted  between  AB  and  Y  at  B. 
But  let  AB  be  considered  as  made  up  of  two  parts,  AC  and  CB. 
Then  (Art.  60)  CB  exerts  upon  AC  a  force  upward,  and  AC 
exerts  upon  CB  a  force  downward.  These  two  forces  are  an 
action  and  its  reaction,  and  constitute  a  stress  which  is  internal 
to  the  body  AB.  This  same  stress  is,  however,  external  to 
either  AC  or  CB.  An  equivalent  stress  evidently  exists  at 
every  section  between  A  and  B.  (When  we  refer  to  the  force 
acting  upon  CB  at  C,  we  mean  the  resultant  of  all  forces  exerted 
upon  the  particles  of  CB  by  the  particles  of  AC.  This  resultant 
is  made  up  of  very  many  forces  acting  between  the  particles. 
Also  the  stress  at  C  means  the  stress  made  up  of  the  two 
resultants  of  the  forces  exerted  by  AC  and  CB  upon  each  other.) 

65.  Three  Kinds  of  Internal  Stress. —  It  is  evident  that  the 
internal  stress  at  C  in  the  body  AB  (Fig.  20)  depends  upon 
the  external  forces  applied  to  the  body.  If  the  forces  at  A  and 
B  cease  to  act,  the  forces  exerted  by  AC  and  CB  upon  each 
other  become  zero.  If  the  forces  at  A  and  B  are  reversed  in 
direction,  so  also  are  those  at  C.  (As  a  matter  of  fact,  the 
particles  of  AC  exert  forces  upon  those  of  CB,  even  if  the 
external  forces  do  not  act.  But  if  the  external  forces  applied 
to  CB  are  balanced,  the  resultant  of  the  forces  exerted  on  CB 
by  AC  is  zero.) 


44 


GRAPHIC   STATICS. 


The  nature  of  the  internal  stress  at  any  point  in  a  body  is 
thus  seen  to  depend  upon  the  external  forces  applied  to  the  body. 

Now,  if  we  consider  two   adjacent   portions  of  a   body  (as 

the  parts  X  and   F,  Fig.  22)  separated  by  a  plane  surface,  the 

external  forces  may  have  either  of  three 

i^  r"j^  (\  tendencies:  (i)  to  pull  X  and  Y  apart 
in  a  direction  perpendicular  to  the  plane 
of  separation ;  (2)  to  push  them  together 

in  a  similar  direction ;  (3)  to  slide  each  over  the  other  along  the 
plane  of  separation.  Corresponding  to  these  three  tendencies, 
the  stress  between  X  and  F  may  be  of  either  of  three  kinds  : 
tensile,  compressive,  or  shearing. 

A  tensile  stress  is  such  as  comes  into  action  to  resist  a 
tendency  of  the  two  portions  of  the  body  to  be  pulled  apart  in 
the  direction  of  the  normal  to  their  surface  of  separation. 

A  compressive  stress  is  such  as  comes  into  action  to  resist  a 
tendency  of  X  and  F  to  move  toward  each  other  along  the 
normal  to  the  surface. 

A  shearing  stress  is  such  as  acts  to  resist  a  tendency  of  X 
and  Fto  slide  over  each  other  along  the  surface  between  them. 

In  case  of  a  tensile  stress,  the  force  exerted  by  X  upon  F 
has  the  direction  from  F  toward  X ;  and  the  force  exerted  by 
Fupon  X  has  the  direction  from  X  toward  F. 

In  case  of  a  compressive  stress,  the  force  exerted  by  X  upon 
Fhas  the  direction  from  ^f  toward  F;  and  the  force  exerted  by 
Fupon  X  has  the  direction  from  Fto  ward  X. 

In  the  case  of  a  shearing  stress,  the  force  exerted  by  X 
upon  Fmay  have  any  direction  in  the  plane  of  separation;  the 
force  exerted  by  F  upon  X  having  the  opposite  direction. 

If  X  and  F  are  separate  bodies,  instead  of  parts  of  one 
body,  a  similar  classification  may  be  made  of  the  kinds  of 
stress  between  them  ;  but  with  these  we  shall  have  no  occasion 
to  deal.  The  terms  tensile  stress,  compressive  stress,  and  shear- 
ing stress  (or  tension,  compression,  and  shear)  are  usually  applied 
only  to  internal  stresses. 


EXTERNAL  AND   INTERNAL   STRESSES. 


45 


66.  Strain.  —  In  what  has  preceded,  the  bodies  dealt  with 
have  been  regarded  as  rigid ;  that  is,  the  relative  positions  of 
the  particles  of  any  body  have  been  regarded  as  remaining 
unchanged.  But,  as  remarked  heretofore,  no  known  body  is 
perfectly  rigid.  If  no  external  forces  act  upon  a  body,  its 
particles  take  certain  positions  relative  to  each  other,  and  the 
body  has  what  is  called  its  natural  shape  and  size.  If  external 
forces  are  applied,  the  shape  and  size  will  generally  be  changed  ; 
the  body  is  then  said  to  be  in  a  state  of  strain.  The  deforma- 
tion produced  by  any  system  of  applied  forces  is  called  the 
strain  due  to  those  forces.  The  nature  of  this  strain  in  any 
case  depends  upon  the  way  in  which  the  forces  are  applied.  It 
is  unnecessary  to  treat  this  subject  further  at  this  point,  since 
we  shall  at  present  be  concerned  only  with  problems  in  the 
treatment  of  which  it  will  be  sufficiently  correct  to  regard  the 
bodies  as  rigid. 

[NOTE.  —  There  is  a  lack  of  uniformity  among  writers  in  regard  to  the  meanings 
attached  to  the  words  stress  and  strain.  It  may,  therefore,  be  well  to  explain  again 
at  this  point  the  way  in  which  these  words  are  used  in  the  following  pages.  The 
word  stress  should  be  employed  only  in  the  sense  above  defined,  as  consisting  of  two 
equal  and  opposite  forces  constituting  an  action  and  its  reaction.  The  two  forces  are 
exerted  respectively  by  two  bodies  or  portions  of  matter  upon  each  other.  An 
internal  stress  is  a  stress  between  two  parts  of  the  same  body.  An  internal  force  is 
one  of  the  forces  of  an  internal  stress.  It  is  intended  in  what  follows  to  use  the 
words  "  internal  stress  "  (or  simply  "  stress  ")  only  when  both  the  constituent  forces 
are  referred  to;  and  when  only  one  of  the  forces  is  meant,  to  use  the  words  "inter- 
nal force"  (or  simply  "force").  It  will  be  noticed,  therefore,  that  in  the  following 
pages  the  words  "force  in  a  bar  "  are  frequently  used  where  many  writers  would  say 
"  stress."  This  departure  from  the  usage  of  many  high  authorities  seems  justified  by 
the  following  considerations :  (i)  It  agrees  with  the  usage  which  is  being  adopted 
by  the  highest  authorities  in  pure  mechanics.  (2)  It  is  desirable  that  the  nomencla- 
ture of  technical  mechanics  shall  agree  with  that  of  pure  mechanics,  so  far  as  they 
deal  with  the  same  conceptions.  The  definition  of  strain  above  given  is  in  conform- 
ity with  the  usage  of  the  majority  of  the  more  recent  text-books.  But  it  is  not  rare 
to  find  in  technical  literature  the  word  strain  used  in  the  sense  of  internal  stress  as 
above  defined.  Such  use  of  the  word  should  be  avoided.] 


i  J 

^    P* 


46  GRAPHIC   STATICS. 

§  3.    Determination  of  Internal  Stresses. 

67.  General  Method.  —  The  stresses  exerted  between  the 
parts  of  a  body  may  or  may  not  be  completely  determinate  by 
means  of  the  principles  already  deduced.  But  in  all  cases 
these  principles  suffice  for  their  partial  determination.  The 
general  method  employed  is  always  the  same,  and  will  now  be 
illustrated.  As  heretofore  we  deal  only  with  complanar  forces. 
Let  XY  (Fig.  23)  represent  a  body  in  equilibrium  under  the 
action  of  any  known  external  forces  as  shown.  Now  conceive 
the  body  to  be  divided  into  two  parts  as  X  and  Y,  separated 
by  any  surface.  The  particles  of  X  near  the  surface  exert 
upon  those  of  Y  certain  forces,  and  are  in  return  acted 

upon  by  forces  exerted  by 
the  particles  of  Y.  These 
forces  are  internal  as  re- 
gards the  whole  body.  In 
/p  P6  order  to  determine  them  so 

'3  Fig.  23 

far  as  possible,  we  proceed 

as  follows  :  Let  the  resultant  of  all  the  forces  exerted  by  Y 
upon  Jf  be  called  T;  then  7~is  either  a  single  force  or  a  couple. 
Now  apply  the  conditions  of  equilibrium  to  the  body  X.  The 
external  forces  acting  on  X  are  Plt  P2,  P3,  and  T.  Since  Plt  P2, 
and  P3  are  supposed  known,  T  can  be  determined.  In  fact, 
T  is  equal  and  opposite  to  the  resultant  of  Plt  P.2,  and  P3. 

So  much  can  always  be  determined.  But  T  is  the  resultant 
of  a  great  number  of  forces  acting  on  the  various  particles 
of  X;  and  these  separate  forces  cannot  in  general  be  deter- 
mined by  methods  which  lie  within  the  scope  of  this  work. 

The  general  principle  just  illustrated  may  be  stated  as 
follows  : 

If  a  body  in  equilibrium  under  any  external  forces  be  conceived 
as  made  up  of  two  parts  X  and  Y,  then  the  internal  forces 
exerted  by  X  upon  Y,  together  with  the  external  forces  acting  on 
Y,  form  a  system  in  equilibrium. 


DETERMINATION   OF   INTERNAL   STRESSES.  47 

As  an  immediate  consequence,  we  may  state  that  tJie  result- 
ant of  the  forces  exerted  by  X  upon  Y  is  equivalent  to  the  result- 
ant of  the  external  forces  acting  on  X ;  and  is  eqiial  and  opposite 
to  the  resultant  of  the  external  forces  acting  upon  Y. 

Example.  —  Assume  a  bar  of  known  dimensions,  and  the 
magnitudes,  directions,  and  points  of  application  of  five  forces 
acting  on  it.  Then  (i)  determine  a  sixth  force  which  will 
produce  equilibrium  ;  and  (2)  assume  the  bar  divided  into  two 
parts  and  find  the  resultant  of  the  forces  exerted  by  each  part 
on  the  other. 

68.  Jointed  Frame.  —  In  certain  ideal  cases  (corresponding 
more  or  less  closely  to  actual  cases),  the  internal  forces  may  be 
more  completely  determined.  The  most  important  of  these 
cases  is  that  which  will  be  now  considered.  Conceive  a  rigid 
body  made  up  of  straight  rigid  bars  hinged  together  at  the  ends. 
Assume  the  following  conditions  : 

(1)  The  hinges  are  without  friction. 

(2)  All   external  forces  acting  on   the  body  are  applied  at 
points  where  the  bars  are  joined  together. 

The  meaning  of  these  conditions  will  be  seen  by  reference  to 
Fig.  24.  The  three  bars  X,  Y,  and  Z  are  connected  by  a  "  pin 
joint,"  the  end  of  each  bar  having  a  hole 
or  "eye"  into  which  is  fitted  a  pin. 
(Of  course  the  three  bars  cannot  be  in 
the  same  plane,  but  they  may  be  nearly 
so,  and  will  be  so  assumed  in  what  fol- 
lows.) Condition  (i)  is  satisfied  if  the 
pin  is  assumed  frictionless.  The  effect  of  this  is  that  the  force 
exerted  upon  the  pin  by  any  bar  (and  the  equal  and  opposite 
reaction  exerted  upon  the  bar  by  the  pin)  act  in  the  normal 
to  the  surfaces  of  these  bodies  at  the  point  of  contact ;  and, 
therefore,  through  the  centers  of  the  pin  and  the  hole.  Condi- 
tion (2)  means  that  any  external  force  (that  is,  external  to  the 


48 


GRAPHIC   STATICS. 


whole  body)  applied  to  any  bar  is  applied  to  the  end  and  in  a 
line  through  the  center  of  the  pin. 

With  the  connection  as  shown,  the  bars  do  not  exert  forces 
upon  each  other  directly.  But  each  exerts  a  force  upon  the 
pin,  and  any  force  exerted  by  Y  or  Z  upon  the  pin  causes  an 
equal  force  to  be  exerted  upon  X.  (This  is  seen  by  applying 
the  condition  of  equilibrium  to  the  pin.)  Hence,  in  considering 
the  forces  acting  upon  any  bar  as  X,  we  may  disregard  the  pin 
and  assume  that  each  of  the  other  bars  acts  directly  upon  X. 
By  what  has  been  said,  all  such  forces  exerted  upon  X  by  the 
other  bars  meeting  it  at  the  joint  may  be  regarded  as  acting  at 
the  same  point  —  the  center  of  the  pin.  We  therefore  treat  the 
bars  as  mere  "  material  lines,"  and  regard  all  forces  exerted  on 
any  bar  (whether  by  the  other  bars  or  by  outside  bodies)  as 
applied  at  the  ends  of  this  "material  line." 

Since,  with  these  assumptions,  all  forces  acting  on  any  bar 
MN  (Fig.  25)  are  applied  either  at  M  or  N,  the  forces  applied 
at  M  must  balance  those  applied  at  N.  The  resultants  of  the 
two  sets  must  therefore  be  equal  and  opposite,  and  have  the 
same  line  of  action  —  namely  MN.  Further,  it  follows  that 
the  stress  in  the  bar,  acting  across  any  plane  perpendicular  to 
its  length,  is  a  direct  tension  or  compression. 

69.  Internal  Stresses  in  a  Jointed  Frame.  —  Let  Fig.  25  rep- 
resent a  jointed  frame  such  as  above  described,  in  equilibrium 
under  any  known  external  forces. 
Let  us  apply  the  general  method 
of  Art.  67  to  this  case.  Divide 
the  body  into  two  parts,  X  and  Y, 
by  the  surface  AB  as  shown.  Now 
apply  the  conditions  of  equilibrium 
to  the  body  X.  The  system  of 
forces  acting  upon  this  body  consists  of  Plt  P2,  P6,  and  the  forces 
exerted  by  Y  upon  X  in  the  three  members  cut  by  the  surface 
AB.  By  Art.  68  the  lines  of  action  of  these  forces  are  known, 


DETERMINATION   OF  INTERNAL   STRESSES.  49 

being  coincident  with  the  axes  of  the  members  cut.  Hence, 
the  system  in  equilibrium  consists  of  six  forces,  three  com- 
pletely known,  and  three  known  only  in  lines  of  action. 
The  determination  of  the  unknown  forces  in  magnitude  and 
direction  is  then  a  case  under  the  general  problem  discussed  in 
Art.  40. 

Nature  of  the  stresses.  —  As  soon  as  the  direction  of  the 
force  acting  upon  X  in  any  one  of  the  members  cut  is  known, 
the  nature  of  the  stress  in  that  member  (whether  tension  or 
compression)  is  known.  For  a  force  toivard  X  denotes  com- 
pression ;  while  a  force  away  from  X  denotes  tension.  (Art. 

65-) 

If  a  section  can  be  taken  cutting  only  two  members,  the 
forces  in  these  may  be  found  by  the  force  polygon  alone.  The 
same  is  true,  if  any  number  of  members  are  cut,  but  the  stresses 
in  all  but  two  are  known. 

The  methods  described  in  the  last  three  articles  will  find 
frequent  application  in  the  chapters  on  roof  and  bridge  trusses, 
Part  II. 

Example.  — Assume  a  jointed  frame  similar  to  the  one  shown 
in  Fig.  25,  and  let  external  forces  act  at  all  the  joints.  Then 
(i)  assume  all  but  three  of  the  forces  known  in  magnitude  and 
direction  and  determine  the  remaining  three  so  as  to  produce 
equilibrium.  (2)  Take  a  section  cutting  three  members  and 
determine  the  stresses  in  those  members. 

70.  Indeterminate  Cases. — If,  in  dividing  the  frame,  more 
than  three  members  are  cut,  the  number  of  unknown  forces  is 
too  great  to  admit  of  the  determination  of  their  magnitudes. 
In  such  a  case,  it  may  happen  that  a  section  elsewhere  through 
the  body  will  cut  but  three  members  ;  and  that  after  the  deter- 
mination of  the  stresses  in  these  three,  another  section  can  be 
taken  cutting  but  three  members  whose  stresses  are  unknown. 
So  long  as  this  can  be  continued,  the  determination  of  the 


50  GRAPHIC   STATICS. 

internal  stresses  can  proceed.  Thus,  in  Fig.  26,  if  a  section 
be  first  taken  at  AB,  there  are  four  unknown  forces  to  be 
determined.  But,  if  the  section  A' B'  be  first  taken,  the 
stresses  in  the  three  members  cut  may  be  determined ;  after 

which   the  section  AB   will 
\A'  introduce  but  three  unknown 

stresses. 

There  may,  however,  be 
cases  in  which  the  stresses 
cannot  all  be  determined  by 
any  method.  With  such  ac- 
tually indeterminate  cases  we  shall  not  usually  have  to  deal.  It 
should  be  noticed,  also,  that  even  when  only  three  members 
are  cut,  the  problem  is  indeterminate  if  these  three  intersect  in 
a  point.  As  in  the  case  just  discussed,  this  indeterminateness 
may  be  either  actual  or  only  apparent ;  in  the  latter  case  it  may 
be  treated  as  above  indicated. 

No  attempt  is  here  made  to  develop  all  methods  that  are 
applicable  or  useful  in  the  determination  of  stresses  in  jointed 
frames.  Some  of  these  are  best  explained  in  connection  with 
the  actual  problems  giving  rise  to  them.  We  have  sought  here 
only  to  explain  and  clearly  illustrate  general  principles. 

71.  Funicular  Polygon  Considered  as  Jointed  Frame.  —  Let 
ab,  be,  cd,  da  (Fig.  27)  be  the  lines  of  action  of  four  forces  in 
equilibrium,  the  force  polygon  being  ABCDA.  Choosing  a 


pole,  draw  any  funicular  polygon,  as  the  one  shown.     Now  let 
the  body  upon  which  the  forces  act  be  replaced  by  a  jointed 


DETERMINATION   OF   INTERNAL   STRES 

frame  whose  bars  coincide  with  the  sides  of  the 
polygon.  If  at  the  joints  of  this  frame  the  given  forces  be 
applied,  the  frame  will  be  in  equilibrium  ;  and  each  bar  will 
sustain  a  tension  or  compression  whose  magnitude  is  repre- 
sented by  the  corresponding  ray  of  the  force  diagram. 

To  prove  this,  we  apply  the  "general  method"  of  Art.  67. 
Consider  any  joint  (as  the  intersection  of  oa  and  ob),  and  let 
the  frame  be  divided  by  a  plane  cutting  these  two  members. 
Then  the  portion  of  the  frame  about  the  joint  is  acted  upon 
by  three  forces :  AB,  acting  in  the  line  ab,  and  forces  acting  in 
the  bars  cut,  their  lines  of  action  being  oa,  ob.  If  the  bar 
oa  sustains  a  compression  and  ob  a  tension,  their  magnitudes 
being  represented  by  OA  and  BO  respectively,  the  portion 
of  the  frame  about  the  joint  will  be  in  equilibrium.  Hence, 
the  tendency  of  the  force  AB  is  to  produce  the  stresses  men- 
tioned in  the  bars  oa,  ob.  In  the  same  way  it  may  be  shown 
that  the  tendency  of  the  force  BC  is  to  produce  in  ob  and 
oc  tensile  stresses  of  magnitudes  OB  and  CO,  respectively. 
Applying  the  same  reasoning  to  each  joint,  it  is  seen  that  every 
part  of  the  frame  will  be  in  equilibrium  if  the  bars  sustain 
stresses  as  follows  :  The  bar  oa  must  sustain  a  compression 
OA  ;  ob  a  tension  OB\  oc  a  tension  OC]  and  od a  compression 
OD.  Hence,  if  the  bars  are  able  to  sustain  these  stresses,  the 
frame  will  be  in  equilibrium. 

If  the  stress  in  any  member  of  the  frame  is  a  tension,  that 
member  may  be  replaced  by  a  flexible  string.  This  is  the 
origin  of  the  name  string  as  applied  to  the  sides  of  the  funic- 
ular polygon.  This  name  is  retained  for  convenience,  but,  as 
just  shown,  it  is  not  always  appropriate. 

72.  Outline  of  Subject. — The  foregoing  pages,  embracing 
Part  I,  have  been  devoted  to  a  development  of  the  principles 
of  pure  statics.  We  pass  next  to  the  application  of  these 
principles  to  special  classes  of  problems. 

Part  II  treats  of   the   determination   of  internal   stresses  in 


52  GRAPHIC   STATICS. 

engineering  structures.  Only  "simple"  structures  are  consid- 
ered, —  that  is,  those  whose  discussion  does  not  involve  the 
theory  of  elasticity.  The  structures  considered  include  roof 
trusses,  beams,  and  bridge  trusses. 

Part  III  develops  the  graphic  methods  of  determining  cen- 
troids  (centers  of  gravity)  and  moments  of  inertia  of  plane 
areas,  including  a  short  discussion  of  "inertia-curves." 


PART  II. 
STX£SS£S  IN  SIMPLE  STRUCTURES. 

•CHAPTER   IV.     INTRODUCTORY. 

§   i.    Outline  of  Principles  and  Methods. 

73.  The  Problem  of  Design.  —  When  any  structure  is  sub- 
jected to  the  action  of  external  forces,  there  are  brought  into 
action  certain  internal  stresses  in  the  several  parts  of  the  struc- 
ture.    The  nature  and  magnitudes  of  these  stresses  depend  upon 
the  external  forces  acting  (Art.  65).     In  designing  the  structure, 
each  part  must  be  so  proportioned  that  the  stresses  induced  in 
it  will  not  become  such  that  the  material  cannot  sustain  them 
without  injury. 

To  determine  these  internal  stresses,  when  the  external  forces 
are  wholly  or  partly  given,  is  the  problem  of  design,  so  far  as 
it  will  be  here  treated. 

74.  External  Forces.  —  The  external  forces  acting  on  a  struc- 
ture must  generally  be  completely  known  before  the  internal 
stresses  can  be  determined.     These  external  forces  are  usually 
only  partly  given,  and  the  first  thing  necessary  is  to  determine 
them  fully. 

The  external  forces  include  (i)  the  loads  which  the  structure 
is  built  to  sustain,  and  (2)  the  reactions  exerted  by  other  bodies 
upon  the  structure  at  the  points  where  it  is  supported.  The 
former  are  known  or  assumed  at  the  outset  and  the  latter  are 
to  be  determined. 

53 


54 


GRAPHIC   STATICS. 


75-  Two  Classes  of  Structures.  —  Structures  may  be  divided 
into  two  classes,  according  as  they  may  or  may  not  be  treated 
as  rigid  bodies  in  determining  the  reactions.  This  may  be 
illustrated  as  follows : 

Let  a  bar  AB  (Fig.  28)  be  supported  in  a  horizontal  position 
at  two  points  A  and  B,  the  supports  being  smooth  so  that  the 
ip  pressures  on  the  bar   at 

•y  — ^  A    and    B    are    vertical. 

W&//M      Let  a  known  load  P  be 


applied  to  the  beam  at  a 
given  point,  and  let  it  be  required  to  determine  the  reactions  at 
A  and  B. 

This  is  a  determinate  problem  ;  for  there  are  three  parallel 
forces  in  equilibrium,  two  being  known  only  in  line  of  action. 
This  problem  was  solved  in  Art.  38. 

But  let  AC  (Fig.  29)  be  a  rigid  bar  supported  at  three  points, 

A,  B,  and   C,  the    reactions   at   those   points   being  vertical. 

IP      ip  ,p  Let  any  known  loads 

J*        i1     i*         B  J 3       C          be   applied   at    given 

points,  and  let  it  be 
required  to  determine 
Fie.  39  the  three  reactions. 

This  problem  is  indeterminate;  for  any  number  of  sets  of 
values  of  the  three  reactions  may  be  found,  which,  with  the 
applied  loads,  would  produce  equilibrium  if  acting  on  a  rigid 
body.  (See  Art.  40.) 

Since,  however,  an  actual  bar  is  not  a  perfectly  rigid  body, 
such  a  problem  as  the  one  just  stated  is,  in  reality,  determinate. 
But  it  cannot  be  solved  without  making  use  of  the  elastic 
properties  of  the  material  of  which  the  body  is  composed. 

The  two  classes  of  problems  are,  therefore,  the  following : 
(i)  those  in  which  the  reactions  can  be  determined  by  treating 
the  structure  as  a  rigid  body,  and  (2)  those  in  which  the 
determination  of  the  reactions  involves  the  theory  of  elasticity. 
We  shall  at  present  deal  only  with  the  former  class  of  prob- 


OUTLINE   OF   PRINCIPLES   AND   METHOD' 

lems.     Structures  coming  under  this  class  will  be  call' 
structures. 

76.  Truss.  —  A  truss  is  a  structure  made  up  of  straight  bars 
with  ends  joined  together  in  such  a  manner  that  the  whole  acts 
as  a  single  body.     The  ends  of  the  bars  are,  in  practice,  joined 
in  various  ways  ;  but  in  determining  the  internal  stresses,  the 
connections  are  assumed  to  be  such  that  no  resistance  is  offered 
by  a  joint  to  the  rotation  of  any  member  about  it.     Such  a 
structure  to  be  indeformable  must  be  made  up  of  triangular 
elements ;   for  more  than  three  bars  hinged  together  in  the 
form    of  a   polygon  cannot  constitute   a   rigid  whole.     If  the 
external  forces  are  applied  to  the  truss  only  at  the  points  where 
the  bars  are  joined,  the  internal  stress  at  any  section  of  a  mem- 
ber will  be  a  simple  tension  or  compression,  directed  parallel  to 
the  length  of  the  bar.     (See  Art.  68.) 

The  most  important  classes  of  trusses  are  roof  trusses  and 
bridge  trusses.  The  methods  used  in  discussing  these  classes 
are,  of  course,  applicable  to  any  framed  structures  under  similar 
conditions. 

77.  Loads  on  a  Truss.  — The  loads  sustained  by  a  truss  may 
be  either  fixed  or  moving.     A  fixed  (or  dead}  load  is  one  whose 
point  of  application  and  direction  remain  constant.     A  moving 
(or  live}  load  is  one  whose  point  of  application  passes  through 
a  series  of  positions.     Fixed  loads  may  be  either  permanent  or 
temporary. 

The  loads  on  a  roof  truss  are  usually  all  fixed,  but  are  of 
various  kinds,  viz.,  the  weight  of  the  truss  itself  and  of  the  roof 
covering,  which  is  a  permanent  load ;  the  weight  of  snow 
lodging  on  the  roof,  and  the  pressure  of  wind,  both  of  which 
are  temporary  loads. 

A  bridge  truss  supports  both  fixed  and  moving  loads.  The 
former  include  the  weight  of  the  truss  itself,  of  the  roadway, 
of  all  lateral  and  auxiliary  bracing  (permanent  loads) ;  and  of 
snow  (a  temporary  load).  The  latter  consist  of  moving  trains 


56  GRAPHIC    STATICS. 

in  the  case  of  railway  bridges,  and  of  teams  or  crowds  of  animals 
or  people  in  the  case  of  highway  bridges. 

78.  Combination  of  Stresses  Due  to  Different  Causes.  —  When 
a  truss  is  subject  to  a  variety   of   external  loads,  it  is  often 
convenient  to  consider  the  effect  of  a  part  of  them  separately. 
If  tensile  and  compressive  stresses   are  distinguished  by  signs 
plus  and  minus,  the  stress  in  any  member  due  to  the  combined 
action  of  any  number  of  loads  is  equal  to  the  algebraic  sum  of 
the  stresses  due  to  the  loads  acting  separately. 

A  proof  of  this  proposition  might  be  given ;  but  it  may  be 
accepted  as  sufficiently  evident  without  formal  demonstration. 

79.  Beams.  —  Another  class    of    bodies    to   be  treated  is 
included  under  the  name  beam. 

A  beam  may  be  defined  as  a  bar  (usually  straight)  resting  on 
supports  and  carrying  loads.  The  loads  and  reactions  are 
commonly  applied  in  a  direction  transverse  to  the  length  of  the 
bar  ;  but  this  is  not  necessarily  the  case. 

The  internal  stresses  in  any  section  of  a  beam  are  less 
simple  than  those  in  the  bars  of  an  ideal  jointed  frame  such  as 
a  truss  is  assumed  to  be.  A  discussion  of  beams  is  given  in 
Chap.  VI. 

80.  Summary  of    Principles   Needed.  —  It  will   be  well   to 
summarize  at  this  point  the  main  principles  and  methods  which 
will  be  employed  in  the  discussion  of  the  problems  that  follow. 

The  general  problem  presented  by  any  structure  consists 
of  two  parts :  (a)  the  determination  of  the  unknown  external 
forces  (or  reactions)  and  (b}  the  determination  of  the  internal 
stresses. 

(a)  In  the  case  of  simple  structures  the  unknown  reactions 
are  usually  two  in  number,  and  the  cases  most  commonly  pre- 
sented are  the  following : 

i st.  — Their  lines  of  action  are  known  and  parallel. 

2nd.  —  The  line  of  action  of  one  and  the  point  of  application 
of  the  other  are  known. 


OUTLINE   OF   PRINCIPLES   AND    METHODS. 


57 


Since  all  the  external  forces  form  a  system  in  equilibrium, 
these  two  cases  fall  under  the  general  problems  discussed  in 
Arts.  38  and  39. 

(b)  In  the  case  of  a  jointed  frame  or  truss,  the  lines  of 
action  of  all  internal  forces  are  known,  since  they  coincide 
with  the  axes  of  the  truss  members.  (Art.  68.)  In  determin- 
ing their  magnitudes  we  may  have  to  deal  with  the  following 
problems  in  equilibrium  : 

ist.  — The  system  in  equilibrium  may  be  completely  known, 
except  the  magnitudes  of  two  forces. 

2nd.  — The  magnitudes  of  three  forces  may  be  unknown. 

The  first  case  may  be  solved  by  simply  making  the  force 
polygon  close.  (See  Art.  35.) 

The  second  case  may  be  solved  by  the  method  of  Art.  40, 
which  consists  in  making  the  force  and  funicular  polygons 
close ;  or  by  the  method  of  Art.  42 ;  or  by  the  principle  of 
moments  (Art.  51). 

The  student  should  be  thoroughly  familiar  with  the  prob- 
lems and  principles  here  referred  to.  In  the  following  chapters 
we  proceed  to  their  application. 

8 1.  Division  of  the  Subject.  —  The  subject  of  the  design  of 
structures,  so  far  as  here  dealt  with,  will  be  treated  in  three 
divisions.  The  first  relates  to  framed  structures  sustaining 
only  stationary  loads ;  the  second  to  beams  sustaining  both 
fixed  and  moving  loads  ;  the  third  to  framed  structures  sustain- 
ing both  fixed  and  moving  loads.  Among  structures  of  the 
first  class,  the  most  important  are  roof  trusses  ;  hence,  these 
are  chiefly  referred  to  in  the  next  chapter.  For  a  similar 
reason,  the  chapter  devoted  to  the  third  class  of  structures 
refers  principally  to  bridge  trusses.  The  chapter  on  beams 
precedes  that  on  bridge  trusses,  for  the  reason  that  the  methods 
used  in  dealing  with  a  beam  under  moving  loads  form  a  useful 
introduction  to  those  employed  in  treating  certain  classes  of 
truss  problems. 


CHAPTER  V.     ROOF  TRUSSES.  —FRAMED  STRUC- 
TURES   SUSTAINING   STATIONARY   LOADS. 

§   i.    Loads  on  Roof  Trusses. 

82.  Weights  of  Trusses.  —  Among  the  loads  to  be  sustained 
by  a  roof  truss  is  the  weight  of  the  truss  itself.  Before  the 
structure  is  designed,  its  weight  is  unknown.  But,  since  it  is 
necessary  to  know  the  weight  in  order  that  the  design  may  be 
correctly  made,  the  method  of  procedure  must  be  as  follows  : 

Make  a  preliminary  estimate  of  the  weight,  basing  it  upon 
knowledge  of  similar  structures  ;  or,  in  the  absence  of  such 
knowledge,  upon  the  best  judgment  available.  Then  design 
the  various  truss  members,  compute  their  weight,  and  com- 
pare the  actual  weight  of  the  truss  with  the  assumed  weight. 
If  the  difference  is  so  great  as  to  materially  affect  the  design  of 
the  truss  members,  a  new  estimate  of  weight  must  be  made, 
and  the  computations  repeated  or  revised.  No  more  than  one 
or  two  such  trials  will  usually  be  needed. 

As  a  guide  in  making  the  preliminary  estimate  -of  weight, 
the  following  formulas  may  be  used.  They  are  taken  from  Mer- 
riman's  "  Roofs  and  Bridges,"  being  intended  to  represent 
approximately  the  data  for  actual  structures,  as  compiled  by 
Ricker  in  his  "  Construction  of  Trussed  Roofs." 

Let  /=span  in  feet  ;  a  =  distance  between  adjacent  trusses  in 
feet  ;  W=  total  weight  of  one  truss  in  pounds.  Then  for 
wooden  trusses 


and  for  wrought  iron  trusses 


LOADS    ON    ROOF   TRUSSES.  (  j,  OF  C. 

83.  Weight  of  Roof  Covering.  —  The  weight  of  roof  co\ 

can  be  correctly  estimated  beforehand  from  the  known  weights" 
of  the  materials.  The  following  data  may  be  employed,  in  the 
absence  of  information  as  to  the  specific  material  to  be  used. 
(See  Merriman's  "  Roofs  and  Bridges,"  p.  4.)  The  numbers 
denote  the  weight  in  pounds  per  square  foot  of  roof  surface. 

Shingling  :  Tin,  I  Ib.  ;  wooden  shingles,  2  to  3  Ibs.  ;  iron,  I 
to  3  Ibs.  ;  slate,  10  Ibs. ;  tiles,  12  to  25  Ibs. 

Sheathing :  Boards  I  in.  thick,  3  to  5  Ibs. 

Rafters:  1.5  to  3  Ibs. 

Purlins  :  Wood,  I  to  3  Ibs.  ;  iron,  2  to  4  Ibs. 

Total  roof  covering,  from  5  to  35  Ibs.  per  square  foot  of  roof 
surface. 

84.  Snow  Loads.  —  The  weight  of  snow  that  may  have  to  be 
borne  will  differ  in  different  localities.     For  different  sections 
of  the  United  States  the  following  may  be  used  as  the  maximum 
snow  loads  likely  to  come  upon  roofs. 

Maximum  for  northern  United  States,  30  Ibs.  per  square  foot 
of  horizontal  area  covered. 

For  latitude  of  New  York  or  Chicago,  20  Ibs*  per  square  foot. 

For  central  latitudes  in  the  United  States,  lolbs.  per  square  foot. 

The  above  weights  are  given  in  Merriman's  "  Roofs  and 
Bridges."  They  are  in  excess  of  those  used  by  some  Bridge 
and  Roof  companies. 

85.  Wind  Pressure  Loads. — The  intensity  of  wind  pressure 
against  any  surface  depends  upon  two  elements  :  (a)  the  velocity 
of  the  wind,  and  (b)  the  angle  between  the   surface  and  the 
direction  of  the  wind. 

Theory  indicates  that  the  intensity  of  wind  pressure  upon  a 
surface  perpendicular  to  the  direction  of  the  wind  should  be 
proportional  to  the  square  of  the  velocity  of  the  wind  relative  to 
the  surface.  As  an  approximate  law  this  is  borne  out  by 
experiment.  If  /  denotes  the  pressure  per  unit  area,  and  v 
the  velocity  of  the  wind,  the  law  is  expressed  by  the  formula 


6o 


GRAPHIC    STATICS. 


/=/hA  Here  k  is  proportional  to  the  density  of  air.  Its 
numerical  value  may  be  taken  as  0.0024,  if  the  units  of  force, 
length,  and  time  are  the  pound,  foot,  and  second  respectively. 

If  the  wind  strikes  a  surface  obliquely,  experiment  shows 
that  the  resulting  pressure  has  a  direction  practically  normal  to 
the  surface.  The  tangential  component  is  inappreciable,  owing 
to  the  very  slight  friction  between  air  and  any  fairly  smooth 
surface.  The  intensity  of  the  normal  pressure  depends  upon 
the  angle  at  which  the  wind  meets  the  surface. 

For  a  given  velocity  of  wind  let  /a  denote  the  normal  pressure 
per  unit  area,  when  the  direction  of  wind  makes  an  angle  a 
with  the  surface,  and  /„  the  pressure  per  unit  area  due  to  the 
same  wind  striking  a  surface  perpendicularly.  Then  the  follow- 
ing formula  *  has  been  given  : 

2  sin  a     . 


It  will  rarely  be  necessary  to  use  values  of  /„  greater  than 
50  Ibs.  per  square  foot.  The  following  table  gives  values  of 
the  coefficient  of  pn  in  the  above  formula  for  different  values 
of  a.  The  value  of  /„  may  be  taken  as  from  40  to  50  Ibs.  per 
square  foot. 


a 

2  sin  a 

a 

2  sin  a 

I  +  sin'2  a 

I  +  sin2  a 

0° 

0.00 

50° 

0.97 

10° 

0-34 

60° 

0.99 

20° 

0.6  1 

70° 

1.  00 

30° 

0.80 

80° 

1.  00 

40°                              0.91 

90- 

1.  00 

*  This  formula  is  given  by  various  writers.  It  is  cited  by  Langley  ("  Experiments  in 
Aerodynamics,"  p.  24),  who  attributes  it  to  Duchemin.  Professor  Langley's  elaborate 
experiments  show  so  close  an  agreement  with  the  formula  that  it  may  be  used  without  hesi- 
tation in  estimating  the  pressure  on  roofs. 


ROOF  TRUSS  WITH  VERTICAL  LOA 

§  2.    Roof  Truss  witJi   Vertical  Loads*. 

86.  Notation.  — The  method  of  determining  internal  stresses 
in  the  case  of  vertical  loading  will  be  explained  by  reference  to 
the  form  of  truss  shown  in  Fig.  30.     The  method  will  be  seen 
to  be  independent  of  the  particular  form  of  the  truss. 

For  designating  the  truss  members  and  the  lines  of  action  of 
external  forces  a  notation  will  be  employed  similar  to  that  used 
in  previous  chapters.  Let  each  of  the  areas  in  the  truss  dia- 
gram be  marked  with  a  letter  or  other  symbol  as  shown  in 
Fig.  30 ;  then  the  truss  member  or  force-line  separating  any 
two  areas  may  be  designated  by  the  two  symbols  belonging  to 
those  areas.  Thus,  the  lines  of  action  of  the  external  forces 
are  ab,  be,  cd,  etc.,  and  the  truss  members  are  gh,  hb,  hi,  etc. 
It  is  to  be  noticed  that  the  lines  representing  the  truss  mem- 
bers represent  also  the  lines  of  action  of  forces, — namely,  the 
internal  forces  in  the  members.  The  joint,  or  point  at  which 
several  members  meet,  may  be  designated  by  naming  all  the 
surrounding  letters.  Thus,  bcih,  hijg  are  two  such  points. 

87.  Loads  and  Reactions. — The  loads  now  considered  are 
assumed  to  be  applied  in  a  vertical  direction,  and  to  act  at  the 
upper  joints  of  the  truss.     This  assumption  as  to  the  points  of    , 
application  may  in  some  cases  represent  very  nearly  the  facts  •-.</ 
in  other  cases  the  loads  will,  in  reality,  be  applied  partly/at 
intermediate  points  on  the  truss  members.     If  the  latter  is  Vhe 
case,  the  load  borne  upon  any  member  is  assumed  to  be  div/ided 
between  the  two  joints  at  its  ends.     In  this  case  the  menSber 
will  be  subject  not  only  to  direct  tension  or  compression,  butl^o 
bending.     With  the  latter  we  are  not  here  concerned,  although 
it  must  always  be  considered  in  designing  the  member. 

The  ends  of  the  truss  are  supposed  to  be  supported  on  hori- 
zontal surfaces,  and  the  reaction  at  each  point  of  support  -^ 

assumed  to  have  a  vertical  direction. 

x, 
If  the  loading  is  symmetrical  with  reference  to  a  vertical  line 


62  GRAPHIC   STATICS. 

through. the  middle  of  the  truss,  it  is  evident  that  each  reaction 
is  equal  to  half  the  total  load.  If  the  loading  is  not  symmetri- 
cal, the  reactions  cannot  be  determined  so  simply.  They  may, 
however,  be  readily  computed  by  either  graphic  or  algebraic 
methods.  Graphically,  the  problem  is  identical  with  that  solved 
in  Art.  38.  The  truss  is  treated  as  a  rigid  body,  the  external 
forces  acting  upon  it  being  the  loads  and  reactions,  which  form 
a  system  of  parallel  forces  in  equilibrium.  Two  of  these  forces 
(the  reactions)  are  unknown  in  magnitude,  but  known  in  line 
of  action.  The  construction  for  determining  their  magnitudes 
is  as  follows  : 

Draw  the  force  polygon  ABCDEF  for  the  five  loads  ;  choose 
a  pole  O,  draw  rays  OA,  OB,  OC,  etc.,  and  draw  the  funicular 
polygon  as  shown  in  Fig.  30.  The  two  polygons  are  to  be 
completed  by  including  the  reactions  FG,  GA,  and  both  poly- 
gons must  close.  We  may  draw  first  og,  the  closing  line  of  the 
funicular  polygon,  and  then  the  ray  OG  parallel  to  it,  thus 
determining  the  point  G  in  the  force  polygon.  The  reactions 
are  now  shown  in  magnitude  and  direction  by  FG  and  GA. 

88.    Determination  of  Internal  Stresses. — When  the  external 

forces  are  all  known,  the  internal  stresses  may  be  found  very 

readily.     The  only  principle  needed  is,  that  for  any  system  of 

forces  in  equilibrium,  the  force  polygon  must  close.     The  con- 

sU^uction  will  now  be  explained. 

V Considering  any  joint  of  the  truss  (Fig.  30)  as  ghijg,  fix  the 
atte  ntion  upon  the  portion  of  the  truss  bounded  by  the  broken 
line;  in  the  figure.  This  portion  is  a  body  in  equilibrium  under 
thre  action  of  four  forces  whose  lines  of  action  coincide  with  the 
a  xes  of  the  four  bars  gh,  hi,  ij,  jg,  respectively.  These  forces 
are  internal  as  regards  the  truss  as  a  whole,  but  external  to  the 
pa:rt  in  question ;  each  force  being  one  of  the  pair  constituting 
l^ie  internal  stress  at  any  point  of  the  bar.  Such  a  force  acts 
.from  the  joint  if  the  stress  in  the  bar  is  a  tension  ;  toward  it  if 
the  stress  is  a  compression.  (Art.  69.) 


ROOF  TRUSS  WITH  VERTICAL  LOADS. 

Since  these  four  forces  form  a  system  in  equilibr 
force  polygon  must  close.     This  condition  will  enable  u 
determine  the  magnitudes  of  the  forces,  provided  all  but 
known,  since  the  polygon  can  then  be  constructed  as  in  Art.  18. 


We  cannot,  however,  begin  with  the  joint  just  considered, 
since  at  first  the  four  forces  are  all  unknown  in  magnitude. 

If,  however,  we  start  with  the  joint  gabh,  the  polygon  of 
forces  can  be  at  once  drawn.  For,  reasoning  as  above,  it  jis 
seen  that  the  portion  of  the  truss  immediately  surrounding 
this  joint  is  in  equilibrium  under  the  action  of  four  forc/es  : 
the  reaction  in  the  line  ga,  the  load  in  the  line  ab,  and,'  the 
internal  forces  in  the  lines  bh,  hg.  Of  these  forces,  two  (the 
reaction  and  the  load)  are  completely  known  ;  and  it  is  neces- 
sary only  to  draw  a  polygon  of  which  two  sides  represent  the 
known  forces  and  the  other  two  sides  are  made  parallel  to  thk 
members  bh,  hg.  Such  a  polygon  is  shown  in  Fig.  30;  and 
BH  and  HG  represent  in  magnitude  and  direction  the  forces 
whose  lines  of  action  are  bh,  hg. 


\ 


64  GRAPHIC   STATICS. 

Evidently,  BH  and  HG  represent  also  the  magnitudes  (Art. 
63)  of  the  internal  stresses  in  the  two  members  bh  and  hg. 
The  nature  of  these  stresses  may  be  found  as  follows  :  Since 
the  four  forces  represented  in  the  polygon  GABHG  are  in 
equilibrium,  and  since  GA,  AB  are  the  directions  of  two  of 
them,  the  directions  of  the  other  two  must  be  BH,  HG. 
Hence,  BH  acts  toivard  the  joint  and  HG  from  it.  This 
shows  that  the  stress  in  bh  is  a  compression,  while  that  in  hg  is 
a  tension. 

Passing  now  to  the  joint  bci/tb,  it  is  seen  that  of  the  four 
forces  whose  lines  of  action  meet  there,  two  are  fully  known, 
namely,  the  load  BC  acting  vertically  downward  and  the  inter- 
nal force  in  hb  acting  toward  the  joint  (since  the  stress  is  com- 
pressive),  while  the  remaining  two  (viz.,  the  internal  forces  in 
ci  and  ih)  are  unknown  in  magnitude  and  direction.  Since, 
however,  the  unknown  forces  are  but  two  in  number,  the  force 
polygon  can  be  completely  drawn,  and  is  represented  by  the 
quadrilateral  HBCIH.  The  directions  of  the  forces  are  found 
as  in  the  preceding  case,  and  it  is  seen  that  the  bars  ci  and  ih 
both  sustain  compressive  stresses. 

The  process  may  be  continued  by  passing  to  the  remaining 
joints  in  succession,  in  such  order  that  at  each  there  remain  to 
be  determined  not  more  than  two  forces.  The  complete  con- 
struction is  shown  in  Fig.  30. 

•  >  It  is  evident  that  the  loads  AB  and  EF  might  have  been 
orhitted  without  changing  the  stresses  in  any  of  the  truss 
m'embers.  For  their  omission  would  leave  as  the  complete 
forc*e  polygon  for  external  forces  BCDEGB,  and  the  two 
reactions  would  be  GB  and  EG ;  but  the  force  diagram  would 
be  otherwise  unchanged. 

The  great  convenience  of  the  notation  adopted  is  now  seen.* 


*  This  is  known  as  Bow's  notation.  The  notation  adopted  in  Part  I  involves  the  same 
idea,  but  it  is  not  usually  employed  in  works  on  Graphic  Statics,  though  possessing  very 
evident  advantages.  It  was  suggested  to  the  writer  by  its  use  in  certain  of  Professor 
Eddy's  works. 


ROOF  TRUSS  WITH  VERTICAL  LOADS.        65 

The  line  representing  the  stress  in  any  member  is  designated 
in  the  force  diagram  by  letters  similar  to  those  which  designate 
that  member  in  the  truss  diagram.  The  latter  is  evidently  a 
space  diagram  (Art.  1 1).  The  force  diagram  is  often  called  a 
stress  diagram,  since  it  shows  the  values  of  the  internal  stresses 
in  the  truss  members. 

89.  Reciprocal  Figures.  —  There  are  always  two  ways  of  com- 
pleting the  force  polygon  when  two  of  the  forces  are  known 
only  in  lines  of  action.     (See  Art.  18.)     Either  way  will  give 
correct    results,  but    unless  a  certain   way  be    chosen,   it  will 
become  necessary  to  repeat  certain  lines  in  the  stress  diagram. 
Thus,  if,  in  Fig.  30,  instead  of  GABHG  we  draw  GABH'G, 
the  lines  GH',  H'B  are  not  in  convenient  positions  for  use  in 
the  other  polygons  of  which  they  ought  to  form  sides.     The 
lettering  of  the  diagrams  will  also  be  complicated.     As  an  aid 
in  drawing  the  lines  in  the  most  advantageous  positions,  it  is 
convenient  to  remember  the  fundamental  property  of  figures 
related  in  such  a  way  as  the  force  and  space  diagrams  shown  in 
Fig.  30. 

Such  figures  are  said  to  be  reciprocal  with  regard  to  each 
other.  The  fundamental  property  of  reciprocal  figures  is  that 
for  every  set  of  lines  intersecting  in  a  point  in  either  figure, 
there  is  in  the  other  a  set  of  lines  respectively  parallel  to  them 
and  forming  a  closed  polygon. 

It  is  also  an  aid  to  remember  that  the  order  of  the  sides  in 
any  closed  polygon  in  the  stress  diagram  is  the  same  as  the 
order  of  the  corresponding  lines  in  the  truss  diagram,  if  taken 
consecutively  around  the  joint.  This  usually  enables  us  at 
once  to  draw  the  sides  of  each  force  polygon  in  the  proper; 
order. 

90.  Order  of  External  Forces  in  Force  Polygon.  — It  will  be^ 
observed  that  in  the  case  above  considered,  in  constructing  the<;; 
force  polygon  for  the  loads  and  reactions,  these  forces  hafve 
been  taken  consecutively  in  the  order  in  which  their  points  of 


66 


GRAPHIC    STATICS. 


application  occur  in  the  perimeter  of  the  truss.  This  is  a 
necessary  precaution  in  order  that  the  stress  diagram  and  truss 
diagram  may  be  reciprocal  figures,  so  that  no  line  in  the  former 
need  be  duplicated. 

This  requirement  should  be  especially  noticed  in  such  a  case 
as  that  shown  in  Fig.  31,  in  which  loads  are  applied  at  lower  as 


krH 

1    V.  n       \ 


.<>       I 

»Xl  j 

^  -H-f^o^.^- 


~SSJjs* 


Fig.  31 


well  as  at  upper  joints.  If  the  reactions  are  found  by  the 
method  of  Art.  87,  without  modification,  the  force  polygon  will 
not  show  the  external  forces  in  the  proper  order,  since  the 
known  forces  are  not  applied  at  consecutive  joints  of  the  truss. 
A  new  polygon  should  therefore  be  drawn  after  the  reactions 
have  been  determined. 

If  desirable  (as  in  some  cases  it  may  be)  to  make  use  of  a 
funicular  polygon  in  which  the  external  forces  are  taken  con- 
secutively, this  may  be  drawn  after  the  reactions  are  deter- 
rnined  and  the  new  force  polygon  is  drawn. 

'Jf  a  load  be  applied  at  some  joint  interior  to  the  truss,  as  at 
M  (Fig-  31),  then  in  constructing  the  stress  diagram  it  should 


STRESSES  DUE  TO  WIND  PRESSURE. 


be  assumed  to  act  at  N,  where  its  line  of  action  intBrsectsUfctt&F 
exterior  member  of  the  truss,  and  the  fictitious 
inserted.  The  stresses  in  the  actual  truss  members 
unaffected  by  this  assumption,  and  such  a  device  is  necessary  in 
order  that  the  stress  diagram  may  be  the  true  reciprocal  of  the 
truss  diagram. 

§  3.    Stresses  Due  to    Wind  Pressure. 

91.  Direction  of  Reactions  Due  to  Wind  Pressure.  —  Since 
the  effective  pressure  of  the  wind  has  the  direction  normal  to 
the  surface  of  the  roof  (Art.  85),  it  has  a  horizontal  component 
which  must  be  resisted  by  the  reactions  at  the  supports. 
These  cannot,  therefore,  act  vertically,  as  in  the  case  when 
the  loading  is  vertical.  Their  actual  directions  will  depend 
upon  the  manner  in  which  the  ends  of  the  truss  are  sup- 
ported. 

If  the  ends  of  the  truss  are  immovable,  the  directions  of  the 
reactions  cannot  be  determined,  since  any  one  of  an  infinite 
number  of  pairs  of  forces  acting  at  the  ends  would  produce 
equilibrium.  (The  same  would  be  true  of  the  reactions  due  to 
vertical  loads.)  In  such  a  case  the  usual  assumption  is  one  of 
the  following:  (i)  the  reactions  are  assumed  parallel  to  the 
loads  ;  (2)  the  resolved  parts  of  the  reactions  in  the  horizontal 
direction  are  assumed  equal. 

In  the  case  of  trusses  of  large  span  it  is  not  unusual  to 
support  one  end  of  the  truss  upon  rollers  so  that  it  is  free  to 
move  horizontally,  the  other  end  being  hinged,  or  otherwise 
arranged  to  prevent  both  horizontal  and  vertical  motion.  This 
allows  for  expansion  and  contraction  with  change  of  tempera- 
ture, as  well  as  for  movements  due  to  the  small  distortions  of 
the  truss  under  loads.  With  this  arrangement  the  reaction,  at 
the  end  supported  on  rollers  must  be  vertical  ;  and  since  the 
point  of  application  of  the  other  reaction  is  known,  both  can 
be  fully  determined  by  the  method  described  in  Art.  39. 


68  GRAPHIC   STATICS. 

92.  Determination  of  Reactions.  —  The  methods  of  finding 
reactions  will  now  be  explained  for  the  three  cases  mentioned  in 
the  preceding  article  :  (i)  Assuming  both  reactions  parallel  to  the 
wind  pressure  ;  (2)  assuming  the  horizontal  resolved  parts  of  the 
two  reactions  equal ;  and  (3)  assuming  one  reaction  vertical. 

(1)  The  first  case  needs  no  explanation,  since  it  is  identical 
with   that    described   in    Art.    87,  except    that    the    loads   and 
reactions  have  a  direction  normal  to  one  surface  of  the  roof, 
instead  of  being  vertical.     It  is  to  be  noticed  that  this  assump- ' 
tion  cannot  be  made  if  the   roof  surface  is  curved,  since  the 
lines  of  action  of   the  forces  will   not  be  parallel.     But  since 
the  direction  of  the  resultant  of  the  loads  will  be  known  from 
the  force  polygon,  both  reactions  may  be  assumed  to  act  par- 
allel to  this  resultant,  and  the  construction  made  as  before. 

(2)  In  the  second  case,  let  each  reaction  be  replaced  by  two 
forces   acting   at   the   support,  one   horizontal  and   the   other 
vertical.     The  two  horizontal  forces  are  known  as  soon  as  the 
force   polygon    for  the   loads '  is  drawn,  and  the   two   vertical 
forces  may  be  found  as  in  the  preceding  case,  since  their  lines 
of  action  are  known. 

In  Fig.  32,  let  ab,  be,  cd  be  the  lines  of  action  of  the  wind 
forces.  Let  the  right  reaction  be  considered  as  made  up  of  a 
horizontal  component  acting  in  de  and  a  vertical  component 
acting  in  ef\  and  let  the  left  reaction  be  replaced  by  a  vertical 
component  acting  in  fg  and  a  horizontal  component  acting  in 
ga.  Draw  the  force  polygon  (or  "load-line")  ABCD.  By  th& 
assumption  already  made  GA  and  DE  are  to  be  equal,  and 
their  sum  is  to  equal  the  horizontal  resoled  part  of  AD. 
Through  the  middle  point  of  AD 'draw  a  vertical  line;  its 
intersections  with  horizontal -lines  through  A  and  D  determine 
the  points  G  and  Ey  so  that  the  two  forces  GA  and  DE  become 
known.  The  only  remaining  unknown  forces  are  the  vertical 
forces  EFand  FG.  Choose  a  pole  O,  draw  rays  to  the  points 
G,  A,  B,  C,  D,  E,  and  then  the  corresponding  strings.  Through 
the  points  determined  by  the  intersection  of  og  with_^Jf,  and  oe 


STRESSES   DUE   TO   WIND   PRESSURE. 

with   ef,   draw  the    string   of.     The    corresponding 
from  O  intersects  EG  in  the  point  F,  thus  determini 
FG.     The  reactions  are  now  wholly  known ;   that 
support  being  FA,  and  that  at  the  right  support  DF. 


(3)  For  the  third  case  the  construction  is  shown  in  Fig.  33 
(A)  and  (B},  for  the  two  opposite  directions  of  the  wind.  The 
method  is  identical  with  that  employed  in  Art.  39.  Only  one 
point  of  the  line  of  action  of  the  left  reaction  is  known,  hence 
this  is  taken  as  the  point  of  intersection  of  the  corresponding 
strings  of  the  funicular  polygon.  One  of  these  strings  can  be 
drawn  at  once,  since  the  corresponding  ray  is  known ;  and  the. 
other  is  known  after  the  remaining  strings  have  been  drawn, 
since  it/ must  close  the  polygon.  The  construction  should  be 
carefully  followed  through  by  the  student. 

The  force  polygons  for  the  two  directions  of  the  wind  are 
distinguished  by  the  use  of  O  and  O'  to  designate  the  two  poles. 


70  GRAPHIC    STATICS. 

In  Fig.  33  (A),  ABCDEFGHIA  is  the  force  polygon  for  the 
.case  when  the  wind  is  from  the  right.  Notice  that  the  points 
A,  B,  C,  D,  coincide.  This  means  that  the  loads  AB,  BC,  CD, 
are  each  zero. 


Fig.  33       P 


DABC 


In  diagram  (B},  ABCDEFGHI A  is  the  force  polygon  foi 
the  case  when  the  wind  is  from  the  left.  The  points  £,  F,  G,  J-fj 
coincide,  because  the  loads  EF,  FG,  GH,  are  each  zero. 

93.  Stress  Diagrams  for  Wind  Pressure.  —  When  the  loads 
and  reactions  due  to  wind  pressure  are  known,  the  internal 
stresses  can  be  found  by  drawing  a  stress  diagram,  just  as  in 
the  case  of  vertical  loads.  The  construction  involves  no  new 
principle,  and  will  be  readily  made  by  the  student.  In  Figs. 
33  (A)  and  33  (B]  are  shown  the  diagrams  for  the  two  directions 
of  the  wind. 


MAXIMUM    STRESSES.  71 

'  U.  o*  **• 

The  stresses  in  all  members  of  the  truss  must  be  detelhimed 
for  each  direction  of  the  wind.  If  the  truss  is  symmeti 
with  respect  to  a  vertical  line,  as  is  usually  the  case,  it  may  be 
that  the  same  stress  diagram  will  apply  for  both  directions  of 
wind.  This  will  be  so  if  the  reactions  are  assumed  to  act  as 
in  cases  (i)  and  (2)  of  the  preceding  article.  In  the  case  repre- 
sented in  Fig.  33,  however,  the  hinging  of  one  end  of  the  truss 
destroys  the  symmetry  of  the  two  stress  diagrams,  and  both 
must  be  drawn  in  full. 

§  4.    Maximum  Stresses. 

94.  General  Principles.  —  For  the  purpose  of  designing  any 
truss  member,  it  is  necessary  to  know  the  greatest  stresses  to 
which  it  will  be  subjected  under  any  possible  combination  of 
loads. 

Stresses  are  combined  in  accordance  with  the  principle  stated 
in  Art.  78,  that  the  resultant  stress  in  a  truss  member  due  to 
the  combined  action  of  any  loads  is  equal  to  the  algebraic  sum 
of  the  stresses  due  to  their  separate  action. 

The  method  will  be  illustrated  by  the  solution  of  an  example 
with  numerical  data. 

95.  Problem — Numerical  Data.  —  Let  it  be  required  to  design 
a  wrought  iron  truss  of  40  ft.  span,  of  the  form  shown  in  Fig. 
34  (PI.  I).     Let  12  ft.  be  the  distance  apart  of  trusses,  and  let 
the  loads  be  as  follows  : 

Weight  of  truss,  to  be  assumed  in  accordance  with  the 
formula  of  Art.  82  :  W=\  al  (i  +TV  /).  This  gives  W=  1800  Ibs. 
Assuming  this  to  be  divided  equally  among  the  upper  panels, 
and  that  the  load  for  each  panel  is  borne  equally  by  the  two 
adjacent  joints,  the  load  at  each  of  the  joints  be,  cd,  de  is  450 
Ibs.  The  loads  at  the  end  joints  may  be  neglected,  being  borne 
directly  at  the  supports. 

Weight  of  roof.  —  This  depends  upon  the  materials  used  and 
the  method  of  construction,  but  will  be  taken  as  6  Ibs.  per  sq.  ft. 


72  GRAPHIC   STATICS. 

of  roof  area,  giving  900  Ibs.  as  the  load  at  each  joint.  This, 
also,  is  a  permanent  load.  Total  permanent  load  per  joint, 
1350  Ibs. 

Weight  of  snow.  —  Taking   this   as    15    Ibs.   per   horizontal 
square  foot,  we  find  1800  Ibs.  as  the  load  at  each  joint. 
Wind  pressure.  —  This  is  computed  from  the  formula 


For  this  case  we  put  sin  a  =  ||  =  f  ;  /„  =  40  Ibs.  per  sq.  ft.  ; 
whence  /a=35  Ibs.  per  sq.  ft.  (about).  This  gives  upon  each 
panel  of  the  roof  5250  Ibs.  Then  with  the  wind  from  either 
side,  the  wind  loads  on  that  side  would  be  2625  Ibs.,  5250  Ibs., 
2625  Ibs.  respectively. 

96.  Stress  Diagrams.  —  We  are  now  ready  to  construct  the 
stress  diagrams. 

The  truss  is  shown  (PL  I)  in  Fig.  34  (A).  Fig.  34  (B)  is  the 
stress  diagram  for  permanent  loads.  No  diagram  for  snow 
loads  is  needed,  since  it  would  be  exactly  similar  to  that  for 
permanent  loads.  The  snow  load  at  any  joint  being  four-thirds 
as  great  as  the  permanent  load,  the  stress  in  any  member  due 
to  snow  is  four-thirds  that  due  to  permanent  loads. 

Fig.  34  (C)  shows  the  stress  diagram  for  the  case  of  wind 
blowing  from  the  left.  The  reactions  are  assumed  to  act  in 
lines  parallel  to  the  loads  —  that  is,  normal  to  the  roof.  With 
this  assumption,  no  separate  diagram  is  needed  for  the  case  of 
wind  from  the  right,  since  such  a  diagram  would  be  exactly 
symmetrical  to  Fig.  34  (C).  For  example,  the  stress  in  the 
member  gh  due  to  the  wind  blowing  from  the  right  is  given  by 
the  line  GM  m  Fig.  34  (C). 

97.  Combination  of  Stresses.  —  After  the  stress  diagrams  are 
completed  for  the  various  kinds  of  loads,  the  stresses  should 
be  scaled  from  the  diagrams  and  entered  with  proper  sign  in  a 
table,  as  follows  : 


MAXIMUM   STRESSES. 


73 


Member. 

Permanent 
Load. 

Snow. 

Wind  R. 

Wind  L. 

Max. 

bh 

-4560 

-6080 

—  6270 

-7925 

-  18570 

ci 

-3760 

—  5010 

—  6270 

-7925 

—  16700 

hi 

—  1080 

-1440 

O 

-5250 

-    7770 

ik 

+  1785 

+  2380 

+    55° 

4-6650 

4-  10820 

g* 

+  3735 

+  4980 

+  2600 

+  8800 

+  17520 

g* 

4-2190 

4-  2920 

+  2300 

4-2300 

+    7410 

gm 

+  3735 

+  4980 

4-8800 

4-  2600 

+  17520 

Ik 

+  1785 

+  2380 

4-6650 

+    550 

4-  10820 

ml 

—  1080 

-1440 

-525° 

0 

-    777° 

dl 

-3760 

—  5010 

-7925 

—  6270 

—  16700 

em 

-4560 

-6080 

-7925 

—  6270 

-  18570 

By  combining  the  results,  the  maximum  stress  in  each  mem- 
ber for  any  possible  condition  of  loading  can  be  determined. 
The  possible  combinations  of  loading  are  the  following :  Perma- 
nent load  alone ;  permanent  and  snow  loads ;  permanent  load, 
and  wind  from  either  direction  ;  permanent  and  snow  loads,  and 
wind  from  either  direction.  The  student  will  readily  under- 
stand the  method  of  combining  the  separate  results. 

The  problem  here  solved  relates  to  a  very  simple  form  of 
truss.  With  some  forms  there  may  occur  a  reversal  of  stress 
in  certain  members,  under  different  conditions  of  loading. 

It  is  to  be  noticed  that  in  the  table  the  word  maximum  is 
used  in  its  numerical  sense,  and  has  no  reference  to  the  algebraic 
sign  of  the  stress. 

98.  Examples.  —  In  Fig.  35  (A)  to  (F),  are  shown  several 
forms  of  truss  for  which  the  student  may  draw  stress  diagrams, 
assuming  loads  in  accordance  with  the  data  given  in  Arts.  82 
to  85.  In  determining  reactions  due  to  wind  pressure,  the 


74 


GRAPHIC   STATICS. 


three  assumptions  mentioned  in  Art.  92  should  all  be  used  in 
different  cases,  that  the  student  may  become  familiar  with  the 
principle  of  each. 


(A.) 


§   5.    Cases  Apparently  Indeterminate, 

99.  Failure  of  Usual  Method.  —  In  attempting  to  construct 
the  stress  diagram  by  drawing  the  force  polygon  for  each  joint 
in  succession,  as  in  the  cases  thus  far  treated,  a  difficulty  is 
met  in  certain  forms  of  truss.  It  may  happen  that  after  pro- 
ceeding to  a  certain  point  it  is  impossible  to  select  a  joint  for 
which  the  force  polygon  can  be  completely  drawn,  the  number 
of  unknown  forces  for  every  joint  being  greater  than  two. 

Thus,  in  the  truss  shown  in  Fig.  36,  if  the  stress  diagram  is 
started  in  the  usual  way,  beginning  at  the  left  support,  the  force 
polygons  for  three  joints  may  be  constructed  without  difficulty, 
thus  determining  the  stresses  in  bl,  Ik,  Im,  cm,  mn,  nk.  But 
the  force  polygon  for  cdqpnmc  cannot  be  constructed,  since 
three  forces  are  unknown,  —  namely,  those  in  dq,  qp,  pn.  And 
at  the  joint  knpsk,  the  stresses  in  ///,  ps,  and  sk  are  unknown. 
The  problem,  therefore,  seems  at  this  point  to  become  indeter- 


CASES  APPARENTLY  INDETERMIN 

minate,  since  either  of  the  two  polygons  can 

any  number  of  ways,  so  far  as  the  known  forces  def( 

can  be  -shown,  however,  that  this  ambiguity  is  only  apparent. 

This  may  be  proved  as  follows  : 

Consider  the  portion  of  the  truss  to  the  left  of  the  broken 
line  M'N'.  It  is  in  equilibrium  under  the  action  of  eight 
forces  ;  five  of  these  (four  loads  and  a  reaction)  are  known  ; 
the  remaining  three  are  the  forces  in  er,  rs,  and  sk.  Now,  the 
problem  of  determining  these  three  unknown  forces  is  the  same 
as  that  treated  in  Art.  40.  It  was  there  found  to  be  a  deter- 
minate problem,  unless  the  lines  of  action  of  the  three  unknown 
forces  intersect  in  a  point  or  are  parallel. 

That  the  problem  is  determinate  may  be  seen  also  from  the 
principle  of  moments  (Art.  51).  The  eight  forces  mentioned 
being  in  equilibrium,  the  sum  of  their  moments  is  zero  for  any 
origin  in  their  plane.  Let  the  origin  be  taken  at  the  point  of 
intersection  of  the  lines  of  action  of  two  of  the  unknown  forces, 
as  er,  rs.  Then  from  the  principle  of  moments  we  have  (since 
the  moments  of  the  two  forces  named  are  zero) :  Algebraic  sum 
of  moments  of  loads  and  reaction  to  left  of  section  +  moment 
of  SK=o.  The  only  unknown  quantity  in  this  equation  is  the 
magnitude  of  SK,  which  may,  therefore,  be  determined.  The 
other  unknown  forces  may  be  found  in  a  similar  manner, 
the  origin  of  moments  being  in  each  case  chosen  so  as  to 
eliminate  two  of  the  three  unknown  forces. 

The  whole  problem  of  drawing  the  stress  diagram  is  now 
seen  to  be  determinate.  For,  as  soon  as  the  stress  in  sk  is 
known,  the  force  polygon  for  the  joint  knpsk  contains  but  two 
unknown  sides,  and  can  be  drawn  at  once.  No  further  diffi- 
culty will  be  met. 

100.    Solution    of    Case   of    Failure  —  First    Method. — The 

reasoning  of  the  preceding  article  suggests  two  methods  of 
treating  the  so-called  ambiguous  case.  These  will  now  be 
described. 


76 


GRAPHIC    STATICS. 


The  first  method  is  to  apply  the  construction  of  Art.  40,  as 
follows  :  Referring  to  Fig.  36,  consider  the  equilibrium  of  the 
portion  of  the  truss  to  the  left  of  the  line  M'N'.  The  system 
of  forces  consists  of  those  whose  lines  of  action  are  ka,  ab,  be, 
cd,  de,  er,  rs,  sk.  Let  them  be  taken  in  the  order  named,  and 
draw  the  force  polygon  for  the  known  forces.  (The  reaction 
ka  is  supposed  to  be  already  determined.)  The  known  part  of 
the  force  polygon  is  KABCDE;  the  unknown  part  is  to  be 
marked  ERSK.  Choose  a  pole  O  and  draw  rays  to  K,  A,  B, 
C,  D,  and  E ;  then  draw  the  corresponding  strings  of  the  funic- 
ular polygon.  Remembering  the  method  of  Art.  40,  we  draw 
first  oe,  making  it  pass  through  the  intersection  of  er  and  rs 


(the  reason  for  this  being  that  it  makes  the  string  os  also  pass 
through  that  point} ;  then  draw  in  succession  od,  oc,  ob,  oa,  ok. 
The  string  ok  intersects  sk  in  a  point  through  which  os  must  be 
drawn.  Hence  os  must  join  that  point  with  the  starting  point 
of  the  polygon.  This  completes  the  funicular  polygon,  except 
the  string  or,  which  must  pass  through  the  intersection  of  er 


CASES  APPARENTLY  INDETERMINAT, 


and  rs  in  some  direction  not  yet  known.     This 
necessary  to  the  solution  of  the  problem. 

Since  os  is  now  known,  OS  may  be  drawn  parallel  to  It  f Porn 
O ;  and  by  drawing  a  line  from  K  parallel  to  the  known  direc- 
tion of  KS,  the  point  5  is  determined,  and  KS  becomes 
known. 

To  find  ER  and  RS  it  is  only  necessary  to  draw  from  S  a 
line  parallel  to  rs,  and  from  E  a  line  parallel  to  cr ;  their  inter- 
section gives  the  point  R,  and  the  force  polygon  for  the  system 
of  forces  considered  is  complete.  The  stresses  in  er,  rs,  and  sk 
being  now  known,  the  stress  diagram  may  be  completely  drawn 
by  the  usual  method. 

It  is  interesting  to  notice  that  the  method  just  described 
determines  the  lines  ER,  RS,  SK,  in  their  proper  position  in 
the  complete  stress  diagram.  The  determination  of  ER  and 
RS  by  this  method  is  not  necessary,  since  the  usual  method  of 
drawing  the  stress-diagram  can  be  carried  out  as  soon  as  SK 
is  known. 

The  funicular  polygon  employed  in  the  above  construction, 
so  far  as  it  belongs  to  the  external  forces,  may  coincide  with 
the  corresponding  part  of  the  funicular  polygon  used  in  deter- 
mining the  reactions.  If  this  is  desired,  two  points  must 
be  observed:  (i)  the  string  oe  must  be  the  first  drawn,  and 
must  pass  through  the  intersection  of  er  and  rs,  and  (2)  the 
pole  must  be  so  chosen  that  ok  will  not  be  nearly  parallel 
to  ks. 

It  should  also  be  noticed  that  the  construction  of  the  funic- 
ular polygon  might  begin  with  the  string  ok,  which  should  then 
be  made  to  pass  through  the  intersection  of  rs  and  sk.  The 
student  will  be  able  to  carry  out  this  construction  without 
difficulty. 

101.  Solution  of  Case  of  Failure  —  Second  Method.  —  It  will 
now  be  shown  how  the  apparently  ambiguous  case  can  be 


78  GRAPHIC   STATICS. 

treated  graphically  by  the  principle  of  moments.  Referring 
again  to  Fig.  36,  consider  the  portion  of  the  truss  to  the  left 
of  section  M'N'.  It  is  acted  upon  by  eight  forces,  of  which 
five  (the  loads  and  the  reaction)  are  known,  and  three  (whose 
lines  of  action  are  er,  rs,  sk)  are  unknown.  The  algebraic  sum 
of  the  moments  of  all  these  forces  about  any  origin  must  be 
zero.  Let  the  origin  be  taken  at  the  point  of  intersection  of 
er  and  rs,  so  that  the  moments  of  the  forces  acting  in  these 
lines  are  both  zero  ;  then  the  sum  of  the  moments  of  the  five 
known  forces,  plus  the  moment  of  the  force  acting  in  the  line 
sk,  must  equal  zero.  Now  the  sum  of  the  moments  of  the 
five  known  forces  may  be  found  by  the  method  of  Art.  56. 
Through  the  origin  of  moments  draw  a  line  parallel  to  the 
resultant  of  the  forces  named  (that  is,  a  vertical  line),  and  let 
i  equal  the  length  intercepted  on  it  by  the  strings  oe,  ok.  Then 
the  required  moment  is  —  iff,  where  H  is  the  pole  distance. 
(The  minus  sign  is  given  in  accordance  with  the  convention 
that  left-handed  rotation  shall  be  positive.)  Let  P  =  unknown 
force  in  line  sk,  and  h  its  moment-arm.  For  the  purpose  of 
computing  the  moment,  assume  the  stress  in  sk  to  be  a  tension  ; 
then  the  force  P  acts  toward  the  right  and  its  moment  is  posi- 
tive, the  value  being  +PA. 

Hence,  Ph-Hi=o\ 


From  this  equation  P  may  be  computed.  The  computation 
may  be  made  graphically  as  follows  :  Draw  (Fig.  36)  a  triangle 
WUV,  making  WU=H  (force  units)  and  WV=h  (linear 
units).  Lay  off  WY=i  (linear  units)  and  draw  YX  parallel 
to  VU.  Then  WX  (force  units)  represents  P.  This  is  readily 
seen,  since  from  the  two  similar  triangles  we  have  the  propor- 

tion —  =  -,  which  agrees  with  the  equation  above  deduced. 
H     Ji 

The  computation  is  simplified  if  the  pole  distance  H  is  taken 
equal  to  as  many   force   units    as  h  is  linear  units  ;    or  if   H 


CASES  APPARENTLY  INDETERMINATE. 


79 


is    some   simple   multiple   of    h.     For,  suppose   H=nh\   then 

PJ»!L  =  ni.     lin=i,P  =  i. 

h 
The  stress  in  sk  is  found  to  be  a  tension,  since  P  is  positive. 

Whatever  the  nature  of  the  stress,  it  may  be  assumed  a  ten- 
sion in  writing  the  equation,  and  the  sign  of  the  value  found 
will  show  whether  the  assumption  coincides  with  the  fact. 

102.  Other  Methods  for  Case  of  Failure.  —  In  certain  cases 
the  method  of  treating  the  "  ambiguous  case  "  may  profitably 
be  varied. 

(1)  The   construction  of   Art.    100  may  be  modified  as  fol- 
lows :  Determine  the  resultant  of    the  known  external  forces 
acting  on  the  portion  of  the  truss  to  one  side  of  the  section 
M'N'.     This  resultant  is  in  equilibrium  with  the  three  unknown 
forces  acting  in  the  members  er,  rs,  sk.     Hence  these  forces  can 
be  determined  by  the  special  method  explained  in  Art.  42. 

The  resultant  of  the  five  known  forces  is  represented  in 
magnitude  and  direction  by  KE  in  the  force  polygon ;  and  its 
line  of  action  passes  through  the  intersection  of  the  strings  oe, 
ok  in  the  funicular  polygon.  Since  this  point  of  intersection  is 
likely  to  be  inaccessible,  the  construction  of  Art.  42  cannot  be 
conveniently  applied.  It  may  be  modified  by  using  instead  of 
KE  the  two  forces  KA  (the  reaction  in  the  line  kd)  and  AE 
(the  resultant  of  the  four  loads,  its  line  of  action  being  deter- 
mined by  the  intersection  of  the  strings  oa  and  oe).  First 
determine  forces  in  the  three  lines  er,  rs,  sk,  which  shall  be  in 
equilibrium  with  KA  ;  then  make  a  similar  construction  for 
AE,  and  combine  the  results. 

(2)  It  has  been  proposed  to  employ  the  following  reasoning : 
Remove  the  members/^  and  qr,  and  insert  another  represented 
by  the  broken  line  in  Fig.  36.     Evidently  this  does  not  change 
the   stress  in  the  member  sk,  since  the  forces  acting  on  the 
truss  to  the  left  of  the  section  M'N'  are  unchanged.     But  with 
this  change  the  difficulty  encountered  in  constructing  the  stress 
diagram  by  the  usual  method  is  avoided.     For  when  the  joint 


go  GRAPHIC   STATICS. 

nmcdqpn  is  reached,  the  forces  acting  there  will  be  all  known 
^ex^fpt  -two.  Let  the  stress  diagram  be  drawn  in  the  usual  way 
until  the  stress  in  sk  is  known.  Then  restore  the  original 
bracing  and  repeat  the  construction,  using  the  value  just  deter- 
mined for  SK. 

This  method  is  convenient  whenever  it  is  applicable.  Cases 
may,  however,  arise,  in  which  it  will  fail.  For  instance,  if 
a  load  is  applied  at  the  joint  pqrsp,  the  members  pq  and  qr 
cannot  both  be  omitted,  and  the  method  cannot  be  applied.  In 
such  cases,  one  of  the  methods  explained  in  the  preceding 
articles  may  be  applied. 

Other  methods  might  be  mentioned,  but  the  foregoing  dis- 
cussion of  the  case  will  probably  be  found  sufficient. 

103.  Failing   Case   in  Other   Forms  of  Truss. — The  usual 
method  of  constructing  the  stress  diagram    may  fail  in  other 
forms  of  truss,  though  the  one  above  described  is  the  most 
common.     In  such  a  case,  the  problem  of  rinding  the  stresses 
may  be  really  indeterminate,  or  only  apparently  so.     Whenever 
it  is  possible  to  divide  the  truss  into  two  parts  by  cutting  three 
members  which  are  not  parallel  and  do  not  intersect  in  one 
point,  the  stresses  in  the  three  members  cut  are  determinate  as 
soon  as  all  external  forces  are  known,  and  can  be  found  by 
methods  already  given.     If  more  than  three  members  are  cut, 
the  problem  of  finding  the  stresses  in  them  is  indeterminate, 
unless  all  but  three  of  these  stresses  are  known.     By  remem- 
bering these  principles,  the  determinateness  of  any  given  prob- 
lem may  readily  be  tested.     (See  Art.  70.) 

§  6.    Three-Hinged  Arch. 

104.  Arched  Truss  Defined.  —  If  a  truss  is  so  supported  that 
when   sustaining  vertical  loads  the  reactions  at  the  supports 
have  horizontal  components  directed  toward  the  center  of  the 


THREE-HINGED   ARCH. 

span,  the  truss  becomes  an  arch.     The  only  kind 

shall  here  consider  is   that    consisting  of  two  parti? 

hinged  together  at  the  crown,  and  each  hinged  at  the  point  of 

support. 

Such  a  truss  is  shown  in  Fig.  38,  in  which  the  two  partial 
trusses  are  hinged  to  the  abutments  at  P  and  Q,  and  connected 
by  a  hinge  at  the  point  R.  Since  a  hinge  at  the  support  allows 
the  reaction  of  the  supporting  body  upon  the  truss  to  take  any 
direction  in  the  plane  of  the  truss,  the  directions  of  the  reactions 
at  P  and  Q  are  unknown,  as  is  also  that  of  the  force  exerted  by 
either  partial  truss  upon  the  other  at  the  point  R.  The  problem 
of  determining  the  reactions  may,  therefore,  at  first  sight  seem 
indeterminate.  It  will  be  shown  in  the  next  article  that  it  is 
in  reality  determinate,  and  that  the  only  principles  needed  in 
the  solution  are  such  as  have  been  already  often  applied  in  the 
preceding  chapters.  The  three-hinged  arch  is  indeed  a  "  simple  " 
structure  (Art.  75),  since  the  theory  of  elasticity  is  not  needed 
in  the  determination  of  the  reactions. 

105.  Reactions  Due  to  a  Single  Load. — The  method  of  find- 
ing the  reactions  is  most  clearly  understood  by  considering  the 


(T) 


effect  of  a  single  load  on  either  partial  truss.  Let  a  load  be 
applied  at  5  (Fig.  38)  and  let  all  other  loads  acting  on  either 
portion  of  the  truss  (including  the  weight  of  the  structure)  be 


82  GRAPHIC    STATICS. 

neglected.  Call  the  two  partial  trusses  X  and  Y,  and  consider 
the  part  Y.  The  only  forces  acting  upon  it  are  the  reaction 
exerted  by  the  abutment  at  Q  and  the  force  exerted  at  R  by 
the  truss  X.  These  two  forces,  being  in  equilibrium,  must 
have  the  same  line  of  action,  which  is,  therefore,  the  line  QR. 
Consider  now  the  body  X.  The  external  forces  acting  upon  it 
are  the  load  at  S,  the  reaction  of  the  abutment  at  P,  and  the 
force  exerted  by  Fat  the  point  R.  But  this  last  force  is  equal 
and  opposite  to  the  force  exerted  by  X  upon  Y,  and  its  line  of 
action  is  therefore  QR.  The  three  forces  acting  upon  the 
body  X  being  in  equilibrium,  their  lines  of  action  must  meet  in 
a  point ;  which  point  is  found  by  prolonging  QR  to  meet  the 
line  of  action  of  the  applied  load.  Let  T  be  this  point,  then 
PT  is  the  line  of  action  of  the  reaction  at  P.  The  reactions 
can  now  be  determined  by  drawing  the  triangle  of  forces.  This 
triangle  is  ABC  in  the  figure,  AB  representing  the  load  at  S,  BC 
the  reaction  in  the  line  QR  (also  marked  be},  and  CA  the  reac- 
tion in  the  line  PT  (marked  also  ca).  Evidently  ABC  may  be 
regarded  as  the  polygon  of  external  forces,  either  for  the  partial 
truss  X,  or  for  the  whole  structure  composed  of  X  and  F;  and 
BC  represents  either  the  force  exerted  by  F  upon  X  at  R,  or 
the  force  exerted  upon  F  by  the  supporting  body  at  Q. 

If,  now,  the  structure  be  loaded  at  other  points,  the  reactions 
due  to  each  load  may  be  found  separately;  the  resultant  of 
all  such  separate  reactions  at  either  support  will  be  the  true 
reaction  at  that  support  when  all  the  loads  act  together.  A 
convenient  method  of  applying  these  principles  will  be  given  in 
the  next  article. 

1 06.  Reactions  and  Stresses  Due  to  Any  Vertical  Loads.  —  In 
Fig.  39  is  represented  a  truss  consisting  of  two  parts  supported 
by  hinges  at  P'  and  Q'  and  hinged  together  at  R!.  Consider 
all  vertical  loads  to  be  applied  at  the  upper  joints,  their  lines  of 
action  being  marked  in  the  usual  way.  We  shall  first  explain 
the  construction  for  finding  the  reactions  at  the  supports  ;  after 


THREE-HINGED  ARCH. 

these  are  determined,  the  drawing  of  the  stress 
present  no  difficulty. 

Since  we  shall  sometimes  deal  with  one  of  the  partia 
and  sometimes  with  the  two  considered  as  a  single  body,  it  will 
be  well  at  the  outset  to  specify  the  external  forces  acting  on 
each  of  these  bodies. 


Fig.  39 


(i)  For  the  partial  truss  at  the  left  we  have  five  applied  loads, 
the  reaction  at  R!  (exerted  by  the  other  truss),  and  the  reaction 
at  P'.  The  force  polygon  for  these  forces  will  be  marked  as 
follows:  ABCDEFLA.  (The  meaning  of  the  letters  will  be 
understood  before  the  force  polygon  is  actually  drawn,  by 
reference  to  the  corresponding  letters  in  the  space  diagram.) 


84  GRAPHIC   STATICS. 

(2)  For  the  right-hand    partial  truss  the  external  forces  are 
five  loads  and  two  reactions,   and  the  force  polygon  will  be 
marked  thus  :  FGHIJKLF.     (Notice  that  FL  and  LF  are  equal 
and  opposite  forces,  being  the  "action  and  reaction"  between 
the  two  trusses  at  R' .) 

(3)  For  the  combined  structure  the  external  forces  are  the 
ten  loads  and  the  reactions  at  P'  and  Q'.     (The  action  and 
reaction  at  R'  become  now  internal  forces.)     The  force  polygon 
will  be  marked  thus  :  ABCDEFGHIJKLA. 

Begin  the  construction  by  drawing  the  force  polygon  for  the 
ten  loads  on  the  whole  structure,  lettering  it  as  just  indicated. 
Choose  a  pole  O,  draw  the  rays,  and  then  the  funicular  polygon 
as  far  as  possible.  Now  consider  the  right  partial  truss  as 
unloaded.  The  resultant  of  the  remaining  five  loads  is  repre- 
sented in  magnitude  and  direction  by  AF\  its  line  of  action 
must  pass  through  the  intersection  of  oa  and  of,  and  is  therefore 
the  line  marked  af.  Now,  reasoning  as  in  the  preceding  article, 
we  see  that  the  reaction  at  Q'  must  act  in  the  line  Q'R'.  Let 
Q'R'  intersect  af  in  T,  then  P' T  is  the  line  of  action  of 
the  reaction  at  P'.  Hence  the  complete  force  polygon  for  the 
whole  truss,  when  the  right  half  is  unloaded,  may  be  found  by 
drawing  from  Fa.  line  parallel  to  Q'R' ,  and  from  A  a  line  parallel 
to  P'T',  prolonging  them  till  they  intersect  at  L1 '.  The  reaction 
at  P'  is  L'A,  and  that  at  Q'  is  FL'.  (The  line  of  action  of  the 
latter  is  marked//'.) 

Next,  consider  the  left  half  to  be  without  loads,  the  other 
half  being  loaded.  The  resultant  of  the  five  loads  now  acting  is 
FK,  its  line  of  action  fk  being  drawn  through  the  point  of 
intersection  of  of  and  ok.  The  reactions  at  P'  and  Q1  for  the 
present  case  have  lines  of  action  P'R'  and  Q'T",  found  just 
as  in  the  first  case  of  loading.  These  reactions  are  therefore 
determined  in  magnitude  and  direction  by  drawing  from  K  a 
line  parallel  to  Q'T"  and  from  F  a  line  parallel  to  P'R',  pro- 
longing them  till  they  intersect  at  L".  The  complete  force 
polygon  for  this  case  of  loading  is  therefore  FGHIJKL"F. 


THREE-HINGED   ARCH.  85 

Consider  now  that  both  parts  of  the  truss  are  loaded.  The 
reaction  at  P'  is  the  resultant  of  the  two  partial  reactions  L'A, 
L"F,  and  the  reaction  at  Q'  is  the  resultant  of  the  two  partial 
reactions  FL',  KL".  From  Ln  arid  L'  draw  lines  parallel 
respectively  to  FL'  and  FL",  intersecting  in  L.  Then  L"L  is 
equal  and  parallel  to  FL' ,  and  LL'  is  equal  and  parallel  to  L"F. 
Hence  KL  and  LA  represent  the  resultant  reactions  at  Q'  and 
P'  respectively.  This  completes  the  polygon  of  external  forces 
for  the  whole  truss,  as  well  as  that  for  each  partial  truss. 

The  stress  diagram  can  now  be  drawn  in  the  usual  way, 
beginning  at  the  point  P'.  The  diagram  for  one  partial  truss 
is  shown  in  Fig.  39  (B). 

If  loads  are  applied  at  the  lower  joints  of  the  trusses,  the 
reactions  due  to  these  may  be  found  in  the  same  manner  as  for 
the  upper  loads.  But  before  beginning  the  determination  of 
the  stresses,  the  polygon  must  be  drawn  for  the  external  forces 
taken  in  order  around  the  truss.  (See  Art.  90.) 

107.  Case  of  Symmetrical  Loading.  —  If  the  two  half  trusses 
are  exactly  similar  and  symmetrically  loaded,  the  determination 
of  reactions  and  stresses  is  much  simplified. 

(1)  As  regards  the  reactions,  symmetry  shows  that  the  forces 
exerted  by  the  two  trusses  upon  each  other  at  R'  are  horizontal. 
Hence,  referring  to  Fig.  39,  and  considering  either  half-truss, 
as  that  to  the  left,  the  line  of  action  of  the  reaction  at  P'  may 
be  found  by  drawing  a  horizontal  line  through  R'  and  prolong- 
ing it  to  intersect  af,  the  line  of  action  of  the  resultant  of  all 
loads  on  the  left  truss  ;  the  line  joining  this  point  of  intersection 
with  P'  is  the  required  line  of  action  of  the  reaction  at  the  left 
abutment.     The  two  reactions  are  now  determined  by  drawing 
from  Fa.  horizontal  line  and  from  A  a  line  parallel  to  the  line 
just  determined,  and  prolonging  them  till  they  intersect. 

(2)  As  to  the  stresses,  only  one  partial  truss  need  be  consid- 
ered, since  the  stress   diagrams  for  the  two  portions  will  be 
symmetrical  figures. 


86 


GRAPHIC   STATICS. 


These  principles  might  have  been  employed  in  the  case  dis- 
cussed in  the  preceding  article ;  but  the  method  there  used  is 
applicable  to  cases  in  which  either  the  trusses  or  the  loads  are 
unsymmetrical. 

108.  Wind  Pressure  Diagram.  —  The  diagrams  for  wind 
pressure  will  present  no  difficulty.  The  determination  of  the 
reactions  will,  indeed,  be  simpler  than  in  the  preceding  case, 
since  only  one  partial  truss  will  be  loaded  at  any  one  time,  and 
the  line  of  action  of  one  reaction  is  therefore  known  at  the 
outset.  The  construction  is  shown  in  Fig.  40. 


Fig.  4O. 


In  computing  wind  pressure  loads,  it  will  be  assumed  that 
each  joint  sustains  half  the  pressure  coming  upon  each  of  the 
two  adjacent  panels.  It  will  be  sufficiently  correct  in  comput- 
ing the  pressure  at  any  joint,  to  assume  the  slope  of  the  roof 


THREE-HINGED   ARCH. 

as  that  of  the  tangent  to  the  roof  curve  at  the 
tion  ;  and  the  direction  of  the  wind  load  may  be 
of  the  normal  to  the  roof  curve  at  the  joint.  The  force '  poly-" 
gon  for  the  wind  loads  is  therefore  not  a  straight  line  when  the 
roof  is  curved.  In  Fig.  40,  this  polygon  is  FGHIJK.  The 
reactions  are  to  be  marked  KL,  LF.  (Since  there  are  no  loads 
on  the  other  half-truss,  the  points  ABCDEF  will  coincide.) 
Choosing  a  pole  O,  draw  the  funicular  polygon  as  shown,  and 
determine  the  line  of  action  of  the  resultant  of  all  the  wind 
loads.  This  line  of  action /£  is  drawn  parallel  to  FK,  through 
the  point  of  intersection  of  the  strings  of  and  ok.  Prolong//^ 
to  intersect  P'R'  produced  at  7";  then  Q'T"  is  the  line  of 
action  of  the  reaction  at  Q'.  The  force  polygon  may  now  be 
completed  by  drawing  KL  parallel  to  Q'T11  and  LF  parallel  to 
P'R'.  The  reactions  KL  and  LF  being  thus  determined,  the 
stress  diagram  can  be  drawn  without  difficulty. 

The  stress  diagram  is  drawn  for  both  partial  trusses,  with  the 
result  shown  in  the  figure.  If  the  two  trusses  are  symmetrical, 
the  diagram  for  the  other  direction  of  the  wind  need  not  be 
drawn  ;  the  stresses  for  this  case  being  found  from  the  diagram 
already  drawn.  If,  however,  the  partial  trusses  are  dissimilar,  a 
second  wind-diagram  must  be  drawn. 

It  is  not  necessary  to  draw  separate  space  diagrams  for  verti- 
cal loads  and  wind-forces.  The  constructions  of  Figs.  39  and 
40  have  been  here  kept  wholly  separate,  in  order  that  the 
explanation  may  be  more  easily  followed. 

109.  Check  by  Method  of  Sections.  —  In  case  of  a  truss  of 
long  span,  especially  when  the  members  have  many  different 
directions  and  are  short  compared  with  the  whole  length  of  the 
span,  the  small  errors  made  in  drawing  the  stress  diagram  are 
likely  to  accumulate  so  much  as  to  vitiate  the  results.  Thus, 
in  Fig.  39,  if,  in  drawing  the  stress  diagram,  we  begin  at  P'  and 
pass  from  joint  to  joint,  there  is  no  check  upon  the  correctness 
of  the  work  until  the  point  R'  is  reached.  At  that  point  we 


88  GRAPHIC    STATICS. 

have  a  second  determination  of  the  reaction  exerted  by  each 
half-truss  upon  the  other ;  and  it  is  quite  likely  that  the  two 
values  found  will  not  agree. 

By  a  method  like  that  employed  in  Art.  100  for  the  "inde- 
terminate "  form  of  truss,  we  may  avoid  the  necessity  of  mak- 
ing so  long  a  construction  before  checking  the  results.  In 
Fig.  39  take  a  section  cutting  the  three  members  eg,  qr,  rl, 
and  apply  the  principles  of  equilibrium  to  the  body  at  the 
left  of  the  section.  The  forces  acting  on  this  body  are  LA, 
AB,  BC,  CQ,  QR,  RL.  Choose  a  pole  O'  and  draw  the  funic- 
ular polygon  for  this  system,  making  the  string  o'c  pass  through 
the  point  of  intersection  of  cq  and  qr.  Draw  successively  the 
strings  o'c,  o'b,  o'a,  o'l,  prolonging  the  last  to  intersect  Ir. 
Through  the  point  thus  determined,  draw  o'r,  which  must  also 
pass  through  the  point  of  intersection  of  cq  and  qr.  As  soon  as 
o'r  is  known,  the  corresponding  ray  O'R  can  be  drawn  in  the 
force  diagram,  and  then  by  drawing  from  L  a  line  parallel  to 
Ir,  the  point  R  is  determined.  We  may  now  close  the  force 
polygon,  since  the  directions  of  the  two  remaining  forces  (cq 
and  qr)  are  known. 

The  stresses  in  the  three  members  cut  being  now  known,  we 
have  a  check  on  the  correctness  of  the  construction  of  the 
stress  diagram,  as  soon  as  these  members  are  reached  in  the 
process. 

This  method  will  be  found  of  great  use,  not  only  for  this 
form  of  truss,  but  for  any  truss  containing  many  members. 

§  7.    Counterbracing. 

1 10.  Reversal  of  Stress.  —  If  the  loads  supported  by  a  truss 
are  fixed  in  position  and  unchanging  in  amount,  the  stress  in 
any  member  remains  constant  in  magnitude  and  kind.  But  in 
most  cases  such  are  not  the  conditions,  and  it  may  happen  that 
under  different  combinations  of  loading,  the  stress  in  a  certain 
member  is  sometimes  tension  and  sometimes  compression.  It 


COUNTERBRACING. 

II 

is  often  thought  desirable  to  prevent  such  changes  o* 
the  design  of  the  members  and  their  connections  being 
simplified.     To  accomplish  this  is  the  object  of  counterbracing. 

in.  Counterbracing.  —  Consider  a  truss  such  as  the  one 
shown  in  Fig.  41,  subjected  to  vertical  loads  and  to  wind  pres- 
sure from  either  side.  A  diagonal  member  such  as  xy  may 
sustain  tension  under  vertical  loads  alone,  or  with  the  wind 
from  the  left ;  while  with  the  wind  from  the  right  it  may  sustain 
compression. 

Now  suppose  xy  removed,  and  a  member  represented  by  the 
broken  line  xy1  introduced.  It  may  easily  be  shown  that  any 

M 


system  of  loading  which  would  cause  compression  in  xy  will 
cause  tension  in  this  new  member ;  and  vice  versa.  For,  divide 
the  truss  by  a  section  MN  cutting  xy  and  two  chord  mem- 
bers as  shown,  and  let  L  be  the  point  of  intersection  of  the 
two  chord  members  produced.  The  kind  of  stress  in  xy  or 
the  other  member  (whichever  is  assumed  to  be  present)  may 
be  determined  by  considering  the  system  of  forces  acting  on 
one  portion  of  the  truss,- as  that  to  the  left  of  the  section  MN. 
Let  the  principle  of  moments  be  applied  to  this  system,  the 
origin  being  taken  at  L.  If  the  external  forces  acting  on  the 
portion  of  the  truss  considered  be  such  as  to  tend  to  cause 
right-handed  rotation  about  L,  the  stress  in  xy  must  be  com- 
pression in  order  to  resist  this  tendency  ;  while,  if  xy  be  replaced 
by  xy1,  a  tension  must  exist  in  that  member  to  resist  right- 
handed  rotation  about  L.  Similarly,  a  tendency  of  the  external 
forces  to  produce  left-handed  rotation  about  L  would  be  resisted 
by  a  tension  in  xy  ;  or  by  a  compression  in  the  member  xy' . 
If  the  two  members  act  at  the  same  time,  the  stresses  in 


9o 


GRAPHIC   STATICS. 


them  will  be  indeterminate.  These  stresses  may,  however,  be 
made  determinate  by  the  following  device  : 

Let  the  member  xy  be  so  constructed  that  it  cannot  sustain 
compression.  Then,  whenever  the  external  forces  are  such  as 
to  tend  to  throw  compression  upon  it,  it  ceases  to  act  as  a 
truss-member,  and  the  member  xy1  receives  a  tension  which  is 
determinate. 

If  the  member  xy'  be  constructed  in  the  same  way,  any 
tendency  to  throw  compression  upon  it  causes  it  immediately 
to  cease  to  act,  and  puts  upon  the  member  xy  a  tension  instead. 

A  member  such  as  xy',  constructed  in  the  manner  mentioned, 
is  called  a  counterbrace. 

1 1 2.  Determination  of  Stresses  with  Counterbracing.  —  The 
use  of  counterbracing  adds  somewhat  to  the  labor  of  deter- 
mining the  maximum  stresses,  since  the  members  actually  under 
stress  are  not  always  the  same.  The  method  of  treating  such 
cases  will  be  illustrated  in  the  next  article  by  the  solution  of  an 
example  ;  but  first  the  main  steps  in  the  process  may  be  outlined 
as  follows  : 

(a)  Construct  separate  stress  diagrams  for  vertical  loads  and 
for  wind  in  each  direction,  assuming  the  diagonals  in  all  panels 
to  slope  the  same  way. 

(b}  Determine  by  comparison  of  these  diagrams  in  which  of 
the  diagonal  members  the  resultant  stress  is  ever  liable  to  be  a 
compression.  Draw  in  counters  to  all  such  members. 

(c)  With    these    counterbraces    substituted    for   the    original 
members,  either  draw  new  stress  diagrams,  or  make  the  neces- 
sary additions  to  those  already  drawn.     If  the  latter  method  is 
adopted,  the  added  lines  should  be  inked  in  a  different  color 
from  that  employed  originally.     (In  some  cases,  this  construc- 
tion may  be  unnecessary  on  account  of  symmetry,  as  will  be 
illustrated  in  the  next  article.) 

(d)  Combine  the  separate  stresses  for  maxima  in  the  usual 
way. 


COUNTERBRACING.  9! 

1 13.  Example.  —  In  Fig.  42  (PI.  II)  are  given  stress  diagrams 
for  a  "  bow-string "  roof-truss  in  which  counterbracing  is  em- 
ployed. At  (A)  is  shown  the  truss  or  space  diagram.  The 
span  is  48  ft.  ;  rise  of  top  chord,  16  ft.  ;  rise  of  bottom  chord, 
8  ft.  The  chords  are  arcs  of  parabolas.  The  whole  truss  is 
divided  into  six  panels  by  equidistant  vertical  members.  The 
distance  apart  of  trusses  is  taken  as  12  ft. 

Assume  the  weight  of  the  truss  at  2400  Ibs.,  and  that  of  the 
roof  at  3600  Ibs. ;  this  makes  the  total  permanent  load  1000  Ibs. 
per  panel.  Take  800  Ibs.  as  the  load  at  each  upper  joint,  and 
200  Ibs.  as  the  load  at  each  lower  joint. 

The  snow  load,  computed  at  .15  Ibs.  per  horizontal  square 
foot,  is  about  1440  Ibs.  per  panel. 

Wind  pressure  is  to  be  computed  from  the  formula  of  Art.  85. 

We  now  proceed  to  apply  the  method  outlined  in  the  preced- 
ing article. 

(a)  Assuming  one  set  of  diagonals  present,  we  construct  the 
stress  diagrams  for  the  various  kinds  of  loads. 

Diagram  for  permanent  loads.  —  This  is  shown  at  (B)  Fig. 
42,  which  needs  little  explanation.  It  will  be  noticed  that  the 
stress  in  every  diagonal  member  is  zero.  This  will  always  be 
the  case  if  the  chords  are  parabolic  and  the  vertical  loads  are 
equal  and  spaced  at  equal  horizontal  distances. 

Diagram  for  snow  loads.  —  Fig.  42  (C)  is  the  stress  diagram 
for  snow  loads.  In  this  case,  also,  the  stresses  in  the  diagonals 
are  all  zero. 

Wind  diagrams.  —  At  (D)  and  (E}  are  shown  the  diagrams 
for  the  two  directions  of  the  wind.  The  only  thing  needing 
special  mention  is  the  method  used  in  laying  off  the  force- 
polygon  for  the  wind  loads.  We  first  compute  the  normal  wind 
pressure  on  each  panel  by  the  formula  of  Art.  85.  We  thus 
find,  when  the  wind  is  from  the  right,  the  following  total  pres- 
sures, taking /„  =  40  Ibs.  per  sq.  ft.  : 

On  panel  d,  1650  Ibs.    On  panel  e,  3870  Ibs.    On  panel/,  5480  Ibs. 


92 


GRAPHIC   STATICS. 


In  Fig.  42  (D)  these  are  laid  off  successively  in  their  proper 
directions  to  the  assumed  scale.  Thus  CD',  D'E',  EG  repre- 
sent respectively  1650  Ibs.  normal  to  dg,  3870  Ibs.  normal  to  es, 
and  5480  Ibs.  normal  to  ft.  Now  each  of  these  loads  is  to  be 
equally  divided  between  the  two  adjacent  joints.  Bisect  CD  at 
D,  D'E'  at  E,  and  E'G  at  F\  then  CD,  DE,  EF,  FG  represent 
the  loads  at  the  joints  cd,  de,  ef,  and  fg  respectively.  The 
"load  line"  is  therefore  CDEFG. 

The  reactions  are  assumed  to  be  parallel  to  the  resultant 
load.  With  this  explanation  the  figures  (D)  and  (E)  will  be 
readily  understood 

(b}  Comparison  of  results.  —  The  stresses  due  to  permanent 
loads,  wind  right,  and  wind  left,  are  shown  in  tabular  form  for 
convenience  of  comparison. 


Member. 

Perm.  Load. 

Snow  Load. 

Wind  R. 

Wind  L. 

Max. 

aj 

-6660 

-9680 

-  4250 

-  14070 

-  16340 

bl 

-5350 

-7780 

-  4880 

-  9650 

-  1313° 

en 

-455° 

-6640 

-  6380 

—  6200 

-  11190 

dq 

-4550 

-6640 

-  9500 

-  4250 

-  14050* 

es 

-5350 

-7780 

-  15030 

-  345° 

-  20380* 

fl 

-6660 

-9680 

-  14070 

-  43°o 

-  20730* 

A 

+  5080 

+  7400 

+   780 

+  I3500 

+  12480 

kh. 

+  4680 

+  6820 

+   700 

+  12450 

+  11500 

mhi 

+  447° 

+  6520 

+   1950 

+  7350 

+  10990 

**, 

+  4470 

+  6520 

+  4100 

+  4100 

+  10990* 

rh. 

+  4680 

+  6820 

+  7680 

+  2050 

+  12360* 

ft, 

+  5080 

+  7400 

+  I3500 

+   800 

+  18580* 

jk 

+  1180 

+  1440 

+    ISO 

+  2650 

+  2620 

Im 

+  1180 

+  1440 

-   280 

+  4100 

+  2620 

np 

+  1180 

+  1440 

-   95° 

+  3800 

+  2620* 

qr 

+  1180 

+  1440 

—  1650 

+  2525 

(  +  2620*  ) 
I  -   470*  / 

si 

+  1180 

+  1440 

-  135° 

+  1250 

f  +  2620*  1 
1  -   170*  J 

kl 

o 

O 

+  1300 

—  4600 

+  1300* 

mn 

o 

o 

+  2700 

—  4100 

+  2700* 

pq 

0 

o 

+  4850 

-  3150 

+  4850* 

rs 

o 

o 

+  7080 

-  195° 

+  7080* 

I 

COUNTERBRACING.  93 

It  is  seen  that  the  diagonals  shown  are  all  in  tension  when 
the  wind  is  from  the  right,  and  all  in  compression  when  the 
wind  is  from  the  left.  Therefore  counters  are  needed  in  all 
panels,  and  the  counters  will  all  come  in  action  whenever  the 
wind  is  from  the  left. 

(c)  Stresses  in  count erbraces.  —  It  is  evident  from  symmetry 
that  no  new  diagrams  are  needed  to  determine  the  stresses  in 
the  counterbraces.     In  fact,  the  counterbrace  in  each  panel  is 
situated  symmetrically  to  the  main  diagonal  in  another  panel, 
and  is  subject  to  exactly  equal  stresses. 

(d]  Combination  for  maxima.  —  In  combining  the  results  for 
the  greatest  stresses  in  the  various  members,  it  will  be  assumed 
that  the  greatest  snow  load  and  the  greatest  wind  load  can 
never  act    simultaneously.     For  each   member,  therefore,  the 
stress  due  to  permanent  load  is  to  be  added  to  the  snow  stress 
or  the  wind  stress,  whichever  is  greater.     Again,  in  combining 
the  tabulated  results,  we  consider  only  the  columns  headed  per- 
manent load,  snow  load,  and  wind  right ;  since  whenever  the 
wind  is  from  the  left,  the  other  system  of  diagonals  is  in  action. 
This  gives  the  results  entered  in  the  last  column. 

We  now  notice  the  following  facts : 

(1)  The  results  given  for  the  diagonal  members  are  the  true 
maximum  stresses. 

(2)  The  stress  found  for  each    diagonal  applies   also  to  the 
symmetrically  situated  counterbrace. 

(3)  For  any  other  member,  we  are  to  choose  between  the 
maximum  found  for  that  member  and  the  value  found  for  the 
symmetrically    situated    member.      Thus,    —20730  is  the  true 
maximum  stress  for  both  aj  and/?;  etc.     (The  numbers  denot- 
ing true  maximum  stresses  are  marked  with  a  (*)  in  the  table.) 

It  is  seen  that  the  verticals,  with  one  exception,  may  sustain 
a  reversal  of  stress.  Thus,_/£  and  st  must  be  designed  for  a 
tension  of  2620  Ibs.,  and  also  for  a  compression  of  170  Ibs.  ; 
while  Im  and  qr  are  each  liable  to  2620  Ibs.  tension  and  to  470 
Ibs.  compression. 


CHAPTER   VI.     SIMPLE    BEAMS. 
§    i.    General  Principles. 

1 14.  Classification  of  Beams.  —  A  beam  has  been  defined  in 
Art.  79.     Beams  may  be  treated  in  two  main  classes,  the  basis 
of  classification  being  that  described  in  Art.  75.     These  two 
classes  will  be  called  simple  and  non-simple  beams  respectively. 
The  present  chapter  deals  only  with  simple  beams,  the  defini- 
tion of  which  may  be  stated  as  follows : 

A  simple  beam  is  one  so  supported  that  it  may  be  regarded 
as  a  rigid  body  in  determining  the  reactions. 

A  simple  beam  may  rest  on  two  supports  at  the  ends  ;  or  it 
may  overhang  one  or  both  supports. 

A  cantilever,  is  any  beam  projecting  beyond  its  supports. 
Such  a  beam  may  be  either  simple  or  non-simple. 

A  continuous  beam  is  one  resting  on  more  than  two  supports. 
Such  a  beam  is  non-simple. 

A  beam  may  be  supported  in  several  ways.  It  is  simply 
supported  at  a  point  when  it  rests  against  the  support  so  that 
the  reaction  has  a  fixed  direction.  It  is  constrained  at  a  point 
if  so  held  that  the  tangent  to  the  axis  of  the  beam  at  that 
point  must  maintain  a  fixed  direction.  If  hinged  at  a  support, 
the  reaction  may  have  any  direction.  We  shall  deal  mainly 
with  the  case  of  simple  support. 

In  what  follows  it  will  be  assumed  that  the  beam  rests  in  a 
horizontal  position,  since  this  is  the  usual  case. 

115.  External  Shear,  Resisting  Shear,  and  Shearing  Stress. 
—  The  external  shear  at  any  section  of  a  beam  is  the  algebraic 

94 


GENERAL   PRINCIPLES. 

sum  of  the  external  vertical  forces  acting  on  the  pc 
beam  to  the  left  of  the  section. 

The  resisting  shear  at  any  section  is  the  algebraic  sum  ol  tne 
internal  vertical  forces  in  the  section  acting  on  the  portion  of 
the  beam  to  the  left,  and  exerted  by  the  portion  to  the  right 
of  the  section. 

The  shearing  stress  at  any  section  is  the  stress  which  con- 
sists of  the  internal  vertical  forces  in  the  section,  exerted  by 
the  two  portions  of  the  beam  upon  each  other.  It  consists  of 
the  resisting  shear  and  the  reaction  to  it.  (See  Art.  63.) 

Let  AB  (Fig.  43)  be  a  beam  in  equilibrium  under  the  action 
of  any  external  forces.  At  any  point  in  its  length,  as  C,  con- 
ceive a  plane  to  be  passed  perpen- 

dicular  to  the  axis  of  the  beam,  and  A~C~        ~B 

consider  the  portion  AC,  to  the  left  FI^. 43 

of  the  section.  The  principles  of  equilibrium  apply  to  the  body 
AC,  and  the  external  forces  acting  upon  it  include,  besides 
those  forces  to  the  left  of  C  that  are  external  to  the  whole  bar, 
certain  forces  acting  across  the  section  at  C  that  are  internal  to 
AB,  but  external  to  AC.  (Art.  61.)  These  latter  forces  com- 
prise that  constituent  of  the  internal  stress  between  AC  and 
CB  which  is  exerted  by  CB  upon  AC. 

Represent  by  Fthe  algebraic  sum  of  the  resolved  parts  in 
the  vertical  direction  of  all  forces  acting  on  AB  to  the  left  of 
the  section  at  C,  upward  forces  being  called  positive.  V  is  the 
external  shear  at  the  given  section  as  above  defined. 

Since  the  body  AC  is  in  equilibrium,  condition  (i),  Art.  58, 
requires  that  the  algebraic  sum  of  the  resolved  parts  in  the 
vertical  direction  of  all  forces  acting  on  it  must  equal  zero. 
Hence  the  forces  acting  on  AC  in  the  section  at  C  must  have  a 
vertical  component  equal  to  —  V.  This  vertical  component  is 
called  the  resisting  shear  in  the  given  section.  This  resisting 
shear  is  one  of  the  forces  of  a  stress  of  which  the  other  is  an 
equal  and  opposite  force  exerted  by  A  C  upon  CB.  This  stress 
is  called  the  shearing  stress  in  the  section,  and  is  called  positive 


96  GRAPHIC    STATICS. 

when  it  resists  a  tendency  of  AC  to  move  upward,  and  of  CB 
to  move  downward. 


1 1 6.  Bending  Moment,  Resisting  Moment,  and  Stress  Moment. 
—  The  bending  moment  at  any  section  of  a  beam  is  the  algebraic 
sum  of  the  moments  of  all  the  external  forces  acting  on  the 
portion  of  the  beam  to  the  left  of  the  section  ;  the  origin  of 
moments  being  taken  in  the  section. 

The  resisting  moment  at  any  section  is  the  algebraic  sum  of 
the  moments  of  the  internal  forces  in  the  section  acting  on  the 
portion  of  the  beam  to  the  left,  and  exerted  by  the  portion  to 
the  right  of  the  section  ;  the  origin  of  moments  being  the  same 
as  for  bending  moment. 

The  stress  moment  or  moment  of  internal  stress  at  any  section 
consists  of  the  two  equal  and  opposite  moments  of  the  forces 
exerted  across  the  section  by  the  two  portions  of  the  beam 
upon  each  other. 

Referring  again  to  Fig.  43,  let  us  analyze  further  the  forces  in 
the  section  at  C.  Applying  to  the  body  AC  the  second  condition 
of  equilibrium  ((2)  of  Art.  58),  and  taking  an  origin  at  a  point 
in  the  section,  we  see  that  the  algebraic  sum  of  the  moments 
of  all  the  external  forces  acting  on  the  beam  to  the  left  of  the 
section  plus  the  sum  of  the  moments  of  the  internal  forces  act- 
ing on  AC  in.  the  section  must  equal  zero.  The  former  sum  is 
defined  as  the  bending  moment  at  the  given  section.  Represent 
it  by  M.  The  latter  sum  is  defined  as  the  resisting  moment  at 
the  section,  and  must  be  equal  to  —M,  by  the  above  principle. 

We  have  thus  far  referred  only  to  the  internal  forces  exerted 
by  CB  upon  AC',  but  evidently  the  equal  and  opposite  forces 
exerted  by  AC  upon  CB  have  a  moment  numerically  equal  to 
M.  The  equal  and  opposite  moments  of  the  equal  and  opposite 
forces  of  the  stress  in  the  section  together  constitute  the  stress 
moment  in  the  section. 

If  the  external  forces  applied  to  the  beam  are  all  vertical,  the 
value  of  M  will  be  the  same  at  whatever  point  of  the  section 


GENERAL   PRINCIPLES.  97 

che  origin  is  taken  ;  since  the  arm  of  each  force  will  be  the  same 
foi  all  origins  in  the  same  vertical  line.  If  the  loads  and  reactions 
are  not  all  vertical,  the  value  of  M  will  generally  depend  upon 
what  point  in  the  section  is  taken  as  the  origin  of  moments. 

1 1 7.  Curves  of  Shear  and  Bending  Moment.  —  The  curve  of 
shear  for  a  beam  is  a  curve  whose  abscissas  are  parallel  to  the 
axis  of  the  beam,  and  whose  ordinate  at  any  point  represents 
the  external  shear  at  the  corresponding  section  of  the  beam. 

Let  AB  (Fig.  44)  represent  a  beam  loaded  in  any  manner,  and 
let  A'B'  be  taken  parallel  to  AB.  At  every  point  of  A'B1 
suppose  an  ordinate  drawn  whose  length  shall  represent  the 
external  sh-ear  at  the  corresponding  A  B 

section  of  AB.  The  line  ab,  join- 
ing the  extremities  of  all  these  ordi- 
nates, is  the  shear  curve.  Positive 
values  of  the  shear  may  be  repre- 
sented by  ordinates  drawn  upward 
from  A'B',  and  negative  values  by 
ordinates  drawn  downward.  (Instead  of  drawing  A'B'  parallel 
to  the  beam,  it  may  be  any  other  straight  line  whose  extremi- 
ties are  in  vertical  lines  through  A  and  B.} 

The  curve  of  bending  moment  for  a  beam  is  a  curve  whose 
abscissas  are  parallel  to  the  axis  of  the  beam,  and  whose  ordinate 
at  any  point  represents  the  bending  moment  at  the  correspond- 
ing section  of  the  beam.  Thus  in  Fig.  44,  A"C"B"  may  repre- 
sent the  bending  moment  curve  for  the  beam  AB.  (Evidently 
A"B"  might  be  inclined  to  the  direction  of  AB,  without  destroy- 
ing the  meaning  of  the  curve.)  Positive  and  negative  values  of 
the  bending  moment  will  be  distinguished  by  drawing  the 
ordinates  representing  the  former  upward  and  those  represent- 
ing the  latter  downward  from  the  line  of  reference  A"B". 

1 1 8.  Moment  Curve  a  Funicular  Polygon.  —  If  the  loads  and 
reactions  upon  the  beam  are  all  vertical,  every  funicular  poly- 
gon for  these  forces  taken  consecutively  along  the  beam,  is  a 


98 


GRAPHIC   STATICS. 


curve  of  bending  moments.  Thus,  let  MN  (Fig.  45)  represent 
a  beam  under  vertical  loads,  supported  at  the  ends  by  vertical 
reactions;  Draw  a  funicular  polygon  for  the  loads  and  reactions 
as  shown. 


c\d   N 


Fig 


Now,  by  definition,  the  bending  moment  at  any  section  is 
equal  to  the  moment  of  the  resultant  of  all  external  forces  act- 
ing on  the  beam  to  the  left  of  the  section,  the  origin  of 
moments  being  taken  in  the  section.  By  Art.  56,  this  moment 
can  be  found  by  drawing  through  the  section  a  vertical  line  and 
finding  the  distance  intercepted  on  it  by  the  two  strings  corre- 
sponding to  the  resultant  mentioned  ;  the  product  of  this  inter- 
cept by  the  pole  distance  is  the  required  moment.  Hence,  if 
oe  (Fig.  45)  is  taken  as  axis  of  abscissas,  the  broken  line  made 
up  of  oa,  oby  oc,  od,  is  a  "curve  of  bending  moments." 

119.  Design  of  Beams.  —  The  principles  involved  in  the 
design  of  beams  will  not  be  here  fully  discussed.  Every  prob- 
lem in  design  involves  the  determination  of  shears  and  bend- 
ing moments  throughout  the  beam  ;  and  the  graphic  methods 
of  determining  these  will  alone  be  considered  in  the  following 
articles. 

§  2.  Seam  Sustaining  Fixed  Loads. 

1  20.  Shear  Curve  for  Beam  Supported  at  Ends.  —  Let  MN 
(Fig.  45)  represent  a  beam  supported  at  the  ends  and  sustain- 
ing loads  as  shown.  Draw  the  force  polygon  ABCD,  and  with 
pole  O  draw  the  funicular  polygon.  The  closing  line  is  oe 


BEAM   SUSTAINING   FIXED   LOADS. 

sf      ¥  >•-— "w 

(marked  also  M"N"),  and  OE  drawn  parallel  to  Mm"  fixes  E, 
thus  determining  DE,  EA,  the  reactions  at  the  sum>orts.  • 

Take  M'N'  as  the  axis  of  abscissas  for  the  shear\u» 

The  shear  at  every  section  can  be  at  once  taken 
force    polygon.     For,   remembering   the  definition  of    external 
shear  (Art.  115)  we  have  : 

Shear  at  any  section  between  M  and  P  is  EA  (positive). 

Shear  at  any  section  between  P  and  Q  is  EB  (positive). 

Shear  at  any  section  between  Q  and  R  is  EC  (negative). 

Shear  at  any  section  between  R  and  N  is  ED  (negative). 

Hence  the  shear  curve  is  the  broken  line  drawn  in  the 
figure. 

121.  Moment  Curve  for  Beam   Supported  at  Ends. — As  in 

Art.  1 1 8,  it  is  seen  that  the  funicular  polygon  already  drawn 
(Fig.  45)  is  a  bending  moment  curve  for  the  given  forces.  For 
the  bending  moment  at  any  section  of  the  beam  is  equal  to  the 
corresponding  ordinate  of  this  polygon,  multiplied  by  the  pole 
distance. 

For  simplicity,  it  will  be  well  always  to  choose  the  pole  so 
that  the  pole  distance  represents  some  simple  number  of  force- 
units. 

The  sign  of  the  bending  moment  is  readily  seen  to  be  nega- 
tive everywhere,  according  to  the  convention  already  adopted 
(Art.  47). 

122.  Shear  and  Moment  Curves  for  Overhanging  Beam. — 
Consider  a  beam  such  as  shown  in  Fig.  46,  supported  at  Q  and 
T,  and  sustaining  loads  at  M,  P,  R,  S,  and  N.. 

This  case  may  be  treated  just  as  the  preceding,  care  being 
exercised  to  take  all  the  external  forces  (loads  and  reactions)  in 
order  around  the  beam. 

The  construction  is  shown  in  Fig.  46.  First,  the  reactions  at 
Q  and  T  are  found  as  in  the  preceding  case,  by  drawing  the 
funicular  polygon,  finding  the  closing  line,  and  drawing  OG 
parallel  to  it.  The  reactions  are  FG  and  GA.  Then  the  value 


100 


GRAPHIC   STATICS. 


of  the  external  shear  can  be  found  for  any  section  from  the 
definition,  and  is  always  given  in  magnitude  and  sign  by  a  cer- 
tain portion  of  the  force  polygon  ABCDEFGA.  The  resulting 
curve  is  shown  in  the  figure,  M'N'  being  the  line  of  reference. 


Similarly,  from  the  definition  of  bending  moment,  and  the 
principle  of  Art.  56,  the  bending  moment  at  any  section  is 
equal  to  the  ordinate  of  the  funicular  polygon  multiplied  by  the 
pole  distance.  The  polygon  is  shaded  in  the  figure  to  indicate 
the  ordinates  in  question.  It  is  seen  that  the  bending  moment 
is  positive  at  every  section  of  the  beam. 

123.  Distributed  Loads.  —  In  all  the  cases  thus  far  discussed, 
the  loads  applied  to  the  beam  have  been  considered  as  concen- 
trated at  a  finite  number  of  points.  That  is,  it  has  been 
assumed  that  a  finite  load  is  applied  to  the  beam  at  a  point, 
Such  a  condition  cannot  strictly  be  realized,  every  load  being 
in  fact  distributed  along  a  small  length  of  the  beam.  If  the 
length  along  which  a  load  is  distributed  is  very  small,  no  im- 
portant error  results  from  considering  it  as  applied  at  a  point. 

In  certain  cases  a  beam  may  have  to  sustain  a  load  which  is 
distributed  over  a  considerable  part  of  its  length,  or  over  the 
whole  length.  Such  a  load  may  be  treated  graphically  with 
sufficient  correctness  by  dividing  the  length  into  parts,  and 


BEAM    SUSTAINING    MWl' -LOADS''  '  '      ' 'QI 

assuming  the  whole  load  on  any  part  to  be  concentrated  at  a 
point.  The  smaller  these  parts,  the  more  nearly  correct  will 
the  results  be. 

It  may  be  remarked  byway  of  comparison  that  while  algebraic 
methods  are  most  readily  applicable  to  the  case  of  distributed 
loads,  the  reverse  is  true  of  graphic  methods,  which  are  most 
easily  applied  in  the  very  cases  in  which  analytic  methods 
become  most  perplexing. 

124.  Design  of  Beam  with  Fixed  Loads. — The  above  exam- 
ples are  sufficient  to  explain  the  method  of  treating  any  simple 
beam   under  fixed  loads.     Under  such   loading  the  shear  and 
bending  moment  at  each  section  of  the  beam  remain  unchanged 
in  value,  and  no  further  discussion  is  necessary  as  a  preliminary 
to  the  design  of  the  beam.     We  proceed  next  to  the  case  of 
beams  with  moving  loads. 

§  3.    Beam  Sustaining  Moving  Loads. 

125.  Curves  of  Maximum  Shear  and  Moment.  —  When  a  beam 
sustains  moving  loads,  the  shear  and  moment  at  any  section  do 
not  remain  constant  for  all  positions  of  the  loads.     In  such  a 
case  it  is  the  greatest  shear  or  moment  in  each  section  that  is 
to  be  used  in  designing  the  beam. 

A  curve  of  maximum  shear  is  a  curve  of  which  the  ordinate 
at  each  point  represents  the  greatest  possible  value  of  the  shear 
at  the  corresponding  section  of  the  beam  for  any  position  of  the 
loads. 

A  curve  of  maximum  moment  is  a  curve  whose  ordinate  at 
each  point  represents  the  greatest  possible  value  of  the  bending 
moment  at  the  corresponding  section  of  the  beam  for  any  posi- 
tion of  the  loads. 

In  the  following  articles  will  be  explained  a  method  of  deter- 
mining any  number  of  points  of  the  curves  just  defined,  in  the 
case  of  a  simple  beam  supported  at  the  ends.  The  moving 


I02  GRAPHIC   STATICS. 

load  will  be  taken  to  consist  of  a  series  of  concentrated  loads 
with  lines  of  actions  at  fixed  distances  apart.  An  example  of 
such  a  load  is  the  weight  of  a  locomotive  and  train  ;  the 
points  of  application  of  the  loads  being  always  under  the  several 
wheels. 

On  PL  III  is  represented  a  load-series  consisting  of  two  loco- 
motives followed  by  a  train  whose  weight  is  assumed  to  be 
uniformly  distributed.  The  numbers  given  represent  half  the 
total  weight  of  the  train,  being  the  weight  borne  by  each  of  the 
two  beams  or  trusses  which  sustain  a  single-track  railroad. 

126.  Position  of  Loads  causing  Greatest  Shear  at  a  Given 
Section.  —  In  order  to  determine  the  position  of  a  given  series 
of  loads  which  causes  the  greatest  positive  shear  at  a  given 
section  of  a  beam,  consider  first  the  way  in  which  the  shear  due 
to  a  single  load  varies  as  the  position  of  the  load  changes. 

Let  Al£1  (PI.  Ill)  represent  the  beam,  arid  Fl  the  section 
under  consideration.  A  load  in  any  position  between  Fl  and  Bl 
causes  at  F1  a  positive  shear  equal  in  magnitude  to  the  reaction 
at  A1  ;  a  load  between  A1  and  Fl  causes  at  Fl  a  negative  shear 
equal  to  the  reaction  at  Bv 

Let  AlBl  =  /,  A^  =  /!,  FlBl  =  /2.     A  load  P  on  A^,  dis- 

tant x  from  Alt  causes  at  Fl  a  shear  --  —  ;  as  the  load  moves 

PI 
from  A   to  F  this  shear  varies  from  o  to  —  -  —  1.     A  load  P  on 


l         l 


F^B^  distant  x  from  Blt  causes  at  Fl  a  shear  —  ;  as  the  load 

PI 
moves  from  B1  to  F1  this  shear  varies  from  o  to  —  ?.     These 

results  may  be  represented  graphically  as  follows  : 

Influence  line.  —  From  a  given  reference  line  A^B^  (PI.  Ill 
(O)»  let  ordinates  be  erected  such  that  the  ordinate  at  any  point 
y2  represents  (to  a  convenient  scale)  the  shear  at  F1  due  to  a 
unit  load  at/j.  The  line  joining  the  extremities  of  these  ordi- 
nates is  called  an  "  influence  line  "  for  the  shear  in  question. 


BEAM    SUSTAINING   MOVING   LOADS.        f          IO? 

S      u.or  c. 

The  above  values  of  the  shear  due  to  a  load  P  in  artk  position 
show  that  for  the  portion  of  the  beam  A^  the 
*a  straight  line  A2FS',  FZF3'  being  equal  in  magnitude  to  -1,  and 
being  laid  off  negatively ;  while  for  the  portion  F1£1  the  influ- 
ence line  is  the  straight  line  BzFZt  F%FS  being  equal  to  -?,  and 

being  laid  off  positively.  The  two  lines  AzF3f  and  F3B2  are  seen 
to  be  parallel. 

The  influence  line  shows  at  a  glance  the  effect  of  a  load  in 
any  position  in  producing  shear  at  Fv  Thus,  a  unit  load  at  J^ 
causes  a  positive  shear  equal  to  J%J%,  and  a  load  P  at  J^  causes 
a  positive  shear  equal  to  P  x  J^J^. 

The  resultant  shear  at  Fl  due  to  any  number  of  loads  may  be 
found  by  multiplying  each  load  by  the  corresponding  ordinate 
of  the  influence  line  and  taking  the  algebraic  sum  of  the 
products. 

Consider,  now,  the  effect  of  any  series  of  loads  brought  on 
the  beam  from  the  right.  Every  load  causes  positive  shear  at 
Fl  until  the  foremost  load  reaches  the  section,  and  the  effect  of 
each  load  increases  as  it  approaches  Fv  As  soon  as  any  load 
has  passed  the  section,  its  effect  is  to  cause  negative  shear,  this 
effect  decreasing  as  the  load  recedes  from  the  section. 

It  thus  appears  that,  in  order  that  the  shear  at  a  given  section 
may  have  its  greatest  positive  value,  (a)  there  should  be  little  or 
no  load  to  the  left  of  the  section,  (b)  the  portion  of  the  beam  to 
the  right  of  the  section  should  be  fully  loaded,  and  (c)  the  loads 
at  the  right  should  be  as  near  the  section  as  possible. 

These  rules  are  not  sufficiently  definite  to  serve  as  an  exact 
guide,.  Suppose  the  first  load  to  be  just  at  the  right  of  Flt  and 
consider  the  effect  of  advancing  the  whole  series  of  loads.  As 
the  first  load  passes  the  section  the  shear  diminishes  by  an 
amount  equal  to  that  load.  But  as  the  loads  advance  the  effect 
of  every  load  in  producing  reaction  at  Av  increases  (thus  increas- 
ing the  shear  at  /^),  while  no  further  decrease  occurs  until  the 
second  load  passes  the  section.  The  net  result  of  bringing  the 


IO4 


GRAPHIC    STATICS. 


second  load  up  to  the  section  may  or  may  not  be  to  increase 
the  shear  over  the  value  it  has  when  the  first  load  is  just  at 
the  right  of  Fv  It  is  thus  uncertain,  without  computation, 
whether  the  first  load  or  some  succeeding  load  should  be  brought 
up  to  the  section  in  order  to  cause  the  greatest  positive  shear. 

Instead  of  resorting  to  trials  to  determine  for  which  position 
the  shear  is  greatest,  we  may  apply  a  simple  rule  which  will 
now  be  deduced. 

Let  P1  =  magnitude  of  foremost  load,  P^  =  magnitude  of  sec- 
ond load,  etc.,  W  being  the  total  load  on  the  beam.  Let  /  = 
total  span,  and  m  =  distance  between  Pl  and  Pz.  Let  x  —  dis- 
tance from  BI  to  the  center  of  gravity  of  W  when  Pl  is  at  the 
given  section.  Let  us  compute  the  shear  when  P1  is  just  at 
the  right  of  the  section,  and  then  determine  the  effect  of  moving 
all  loads  to  the  left,  until  the  second  load  comes  to  the  section. 

When  P1  is  at  the  right  of  the  section,  the  shear  is  equal  to 
the  reaction  at  Av  say  R.  Then  (calling  V  the  external  shear) 
we  have 


If  now  the  loads  be  moved  until  Pz  comes  to  a  point  just 
at  the  right  of  the  section,  the  reaction  due  to    W  becomes 

W^x    m\  and  the  shear  at  the  section  becomes 

-pv 


The  increase  of  the  shear  is,  therefore, 


This  increase  is  plus  or  minus  according  as  —  —  is  greater  or 

P  / 
less  than  />,  ;  that  is,  as  Wis  greater  or  less  than  —  L.       Hence 

M 

the  following  rule  : 


BEAM    SUSTAINING   MOVING   LOADS. 


105 


The  maximum  positive  shear  in  any  section  of  the  beam  occurs 

when  the  foremost  load  is  infinitely  near  the  section,  provided  W 

PI  PI 

is  not  greater  than  —  i-  .     If  W  is  greater  than  —  !-,  the  greatest 
m  m 

shear  will  occur  when  some  succeeding  load  is  at  the  section. 

In  the  above  discussion  it  has  been  assumed  that  in  bringing 
P2  up  to  the  section  no  additional  loads  are  brought  upon  the 
beam.  If  this  assumption  is  not  true,  let  W  be  the  new  load 
brought  on  the  beam,  and  x'  the  distance  of  its  center  of 
gravity  from  the  right  support  when  Pz  is  at  the  section.  Then 
the  shear  corresponding  to  this  position  is 


and  the  increase  of  shear  due  to  the  assumed  change  in  the 
position  of  the  load  is 

WmW'x'      „ 


Hence,  in  the  statement  of  the  above  rule,  we  have  only  to 

W/'r' 

substitute    Wm+  W'x'  for   Wm  ;   or,    W+^-^   for   w.     The 

m 

additional  load  W  may  be  neglected  except  when  the  condition 

p  i 
W=  —  —  is  nearly  satisfied  ;  for  the  term  Wx'  will  always  be 

m 

small. 

The  application  of  this  rule  is  very  easy,  and  will  save  much 
labor  in  the  graphic  construction  of  the  shear  curve. 

If  it  is  found  that  the  first  load  should  be  past  the  section 
for  the  position  of  greatest  shear,  we  may  determine  whether 
the  second  or  the  third  should  be  at  the  section  by  an  exactly 
similar  method.  We  have  only  to  apply  the  above  rule,  substi- 
tuting P2  for  Plt  and  for  m  the  distance  between  P2  and  P3. 

127.  Determination  of  Greatest  Shear.  —  Having  determined 
what  position  of  the  load-series  causes  the  greatest  shear  at  a 
given  section,  the  value  of  this  shear  may  be  determined  by 


106  GRAPHIC   STATICS. 

the  method  already  explained  in  the  discussion  of  fixed  loads 


"Draw  in  succession  the  lines  of  action  of  the  loads  at  their 
proper  distances  apart  (1-2,  2-3,  .  .  .,  PL  III).  Draw  the  force 
polygon  *  1-2-3-  •  •  •>  choose  a  pole  O,  and  construct  a  funicular 
polygon  for  the  series  of  loads.  The  same  lines  of  action  may 
be  used  for  all  positions  of  the  load-series,  for  instead  of  moving 
the  loads  in  the  desired  direction,  the  loads  may  be  regarded  as 
fixed  and  the  beam  moved  in  the  opposite  direction. 

At  (5),  PL  III,  is  shown  a  beam  A^B^  of  64  ft.  span,  in  such 
position  that  the  shear  at  the  section  Flt  16  ft.  from  Av  has  its 
greatest  possible  value.  The  second  load  is  at  the  section 
instead  of  the  first,  for  applying  the  test  we  have  P1  =  8ooo  Ibs., 

P  / 

7=64  ft.,  m  =  S.i  ft,  _1-  =63200  Ibs.  ;    while  the  total  load  on 
in 

the  beam  is  104,000  Ibs.  when  the  first  load  is  at  the  section. 

The  beam  being  in  the  position  shown  at  (5),  vertical  lines 
drawn  through  A1  and  Bl  intersecting  the  strings  01  and  02'  at 
A  and  B,  determine  AB,  the  closing  string  of  the  funicular 
polygon  for  the  system  of  forces  acting  upon  the  beam  in  the 
assumed  position.  If  OK  is  the  ray  drawn  parallel  to  AB,  the 
reactions  at  A1  and  Bl  are  Ki  and  2*K  respectively.  The  load 
2-3  being  just  at  the  right  of  the  section  Flt  the  shear  at  that 
section  is  equal  to  the  reaction  at  A±  minus  the  load  1-2  ;  its 
value  is  therefore  K2  in  the  force  polygon. 

The  diagram  of  maximum  positive  shear  is  shown  at  (A), 
PL  III,  the  ordinates  representing  greatest  shear  due  to  the 
moving  loads  being  laid  off  upward  from  A^BV  Thus  the  value 
just  determined  is  laid  off  from  the  point  Fv  The  entire  curve 
is  shown  at  (A),  but  the  construction  is  given  only  for  the 
section  F^ 

The  foregoing  construction  has  referred  only  to  moving  loads. 
The  actual  greatest  shear  at  any  section  is  found  by  combining 

*  The  lines  representing  the  loads  are  for  convenience  drawn  upward  instead  of  down- 
ward. This  does  not,  of  course,  affect  the  construction. 


BEAM   SUSTAINING   MOVING   LOAD 

the  maximum  live-load  shear  with  the  shear  dueto  nermanent 
loads.     Shears  due  to  the  dead  loads  are  re 
PI.  Ill,  being  determined  as  follows. 

128.  Shear  Curve  for  Combined  Fixed  and  Moving  Loads. — 

Let  the  beam  sustain  a  fixed  load  of  25000  Ibs.  uniformly  dis- 
tributed along  the  beam.  The  shear  close  to  the  left  support 
due  to  this  load  is  equal  to  the  reaction,  or  12500  Ibs. ;  and 
decreases  as  we  pass  to  the  right  by  **££&  Ibs.  for  each  foot. 
At  the  middle  of  the  beam  the  shear  is  zero,  and  at  the  right 
support  it  is  —  12500  Ibs.  Hence  the  shear  curve  is  a  straight 
line,  and  may  be  drawn  as  follows :  From  A1  lay  off  an  ordinate 
downward  representing  12500  Ibs.,  and  from  Bl  an  ordinate 
upward  representing  12500  Ibs. ;  the  straight  line  joining  the 
extremities  of  these  ordinates  is  the  shear  curve  for  the  fixed 
load.  Positive  shears  are  laid  off  downward  and  negative 
shears  upward  for  the  reason  that,  if  this  be  done,  the  greatest 
positive  shear  at  any  point  due  to  fixed  and  moving  loads  is 
represented  by  the  total  ordinate  measured  between  the  shear 
curves  for  fixed  loads  and  for  moving  loads.  It  is  seen  that 
at  a  certain  point  somewhere  to  the  right  of  the  center  of  the 
beam  this  greatest  shear  is  zero,  and  for  all  sections  to  the 
right  of  this  point  it  is  negative.  This  point  is  determined  by 
the  intersection  of  the  two  shear  curves. 

129.  Approximate  Position  of  Loads  causing  Greatest  Bending 
Moment  at  a  Given  Section.  —  Let  it  be  required  to  determine  the 
greatest  bending  moment  at  the  section  F1  of  the  beam  A^B^ 
shown   on    PL    III.     Let  A1F1  =  /1,  F^B^=l^  A1B1  =  /1  +  /2=L 
Consider  the  effect  of  a  single  load  in  any  position. 

The  bending  moment  at  F1  due  to  a  loadP  on  A-^F^  distant  x 

PI  x 
from  Alt  is  — j—.     For  a  load/5  on  F^B^  distant  x  from  Blt  the 

bending  moment  at  F1  is  — ^--  The  variation  of  each  of  these 
values  as  the  position  of  the  load  changes  may  be  represented 
graphically  as  follows  : 


108  GRAPHIC   STATICS. 

Influence  line.  —  From  A2B2  ((D),  PL  III),  let  ordinates  be 
erected  such  that  the  length  of  the  ordinate  at  any  point  repre- 
sents, on  a  convenient  scale,  the  bending  moment  at  Fv  due  to 
a  unit  load  at  the  corresponding  point  of  the  beam.  The  line 
joining  the  extremities  of  these  ordinates  is  the  "influence  line" 
for  bending  moments  at  F^. 

At  a  distance  x  from  Az,  the  ordinate  has  the  value  -^-,  which 

// 

varies  from  o  at  A%  to  -^  at  F2.     At  a  distance  x  from  B^  the 

value  of  the  ordinate  is  -j-,  which  varies  from  o  at  B^  to  -~ 
at  Fz.  The  influence  line  therefore  consists  of  the  two  straight 
portions  AZF3,  F3B2,  where  FZF3=^~.  The  ordinates  are  laid 

off  downward,  the  bending  moment  being  always  negative 
according  to  the  convention  already  adopted  (Art.  47  and 
Art.  116). 

The  influence  line  having  been  drawn,  the  bending  moment 
at  Fl  due  to  a  load  at  any  point  Jl  may  be  found  by  multiply- 
ing the  load  by  the  corresponding  ordinate  /2/3-  The  bending 
moment  at  Fl  due  to  the  combined  action  of  several  loads  is 
found  by  multiplying  each  load  by  the  corresponding  ordinate 
of  the  influence  line  and  adding  the  products. 

Since  the  sign  of  the  bending  moment  due  to  a  load  is  the 
same,  whatever  its  position,  and  since  the  bending  moment  due 
to  a  given  load  increases  as  the  load  approaches  the  section,  the 
following  general  principle  may  be  stated  : 

The  bending  moment  at  any  section  has  its  greatest  value 
when  the  beam  is  as  fully  loaded  as  possible,  with  the  heaviest 
loads  near  the  section. 

This  principle  serves  only  as  a  rough  general  guide.  An 
exact  rule  will  be  deduced  in  the  following  Articles. 

130.  Load  Polygon.  —  Let  a  curve  or  broken  line  be  drawn, 
of  which  the  abscissa  is  horizontal,  and  the  ordinate  at  any 
point  represents  (on  the  assumed  force  scale)  the  total  load  up 


BEAM    SUSTAINING   MOVING   LOADS. 


109 


to  that  point,  measured  from  some  fixed  point  in  the  load  series 
Such  a  curve  or  broken  line  may  be  called  a  load  polygon.  In 
the  figure  on  PI.  IV,  Q1  R'  is  a  load  polygon  for  the  series  of 
train  loads  shown  on  PL  III,  QR  being  a  funicular  polygon 
for  the  same  series. 

A  simple  relation  exists  between  the  load  polygon  and  the 
funicular  polygon,  which  is  of  use  in  the  discussion  of  the  prob- 
lem now  before  us. 

Let  x  and  y  be  the  coordinates  of  any  point  of  the  funicular 
polygon,  the  j-axis  (L  F)  being  vertical,  and  the  ;r-axis  any  hori- 
zontal line  LX,  Let  x\  y'  be  the  coordinates  of  the  correspond- 
ing point  of  the  load  polygon,  referred  to  a  vertical  axis  (L'  F'), 
and  a  horizontal  axis  (L'X')  passing  through  the  pole  of  the 
force  polygon  used  in  the  construction  of  the  funicular  polygon. 

Consider  any  point  of  the  funicular  polygon,  for  example  a 

point  on  the  string  02'.     The  value  of  —  at  this  point  is  the 

ax 

tangent  of  the  angle  between  02'  and  the  horizontal.  But  since 
02'  is  parallel  to  the  corresponding  ray  of  the  force  polygon, 

this  tangent  is  equal  to  ^_,  where  H  is  the  pole  distance  used 
H 

in  drawing  the  funicular  polygon.     That  is, 

^=y 

dx    //' 


131.  Variation  of  Bending  Moment  at  Any  Section  of  a  Beam. 
—  Let  A1B1  (at  (A),  PL  IV)  represent  one  position  of  a  beam 
of  span  /,  and  let  F1  be  any  section  at  which  it  is  desired  to 
study  the  variation  of  the  bending  moment.  Let  /1  =  ^1F1, 
/2  =  F1B1,  and  ,3-  =  horizontal  distance  of  A^  from  the  vertical 
axisZ'F'.  Through  Alt  Bv  and  F1  draw  vertical  lines,  inter- 
secting the  funicular  polygon  in  A,  B,  F,  and  the  load  polygon 
in  A',  B',  F1.  Let  the  vertical  through  Fl  intersect  AB  in  G, 
and  A'B'  in  G'  .  Let  the  ordinates  of  A,  B,  F,  G,  measured 


1  10  GRAPHIC   STATICS. 

from  LX),  be  denoted  by  a,  b,  f,  g;  and  the  ordinates  of  A',  B', 
F',g'^  measured  from  L'X',  by  a',  b',f,g>. 

If  M  denotes  the  bending  moment  at  Fv  and  H  the  pole 
distance, 

M=HxFG=H(g-f);  .........  (2) 

and  it  will  now  be  shown  that  the  rate  of  change  of  this  bend- 
ing moment  (i.e.  the  change  per  unit  of  horizontal  displacement) 
as  the  beam  moves  horizontally  while  the  loads  remain  sta- 
tionary is  equal  to  F'  G'  or  g1  —  f. 

Since  AGB  is  a  straight  line,  and  7^=-p  it  follows  that 

LrB       /2 


or 


/ (3) 

*  i> 

In  an  exactly  similar  manner  it  may  be  shown  that 

<^      J       /         /  

From  (2)  and  (3), 


(5) 


If  the  beam  moves  horizontally,  the  abscissas  of  Alt  Blt  and  Fl 

change  at  the  same  rate.     Differentiating  (5)  with  respect  to  z, 


dM         fada     l^M     df\ 
dz          \ldz~*  Idz     dz) 


Now  —,  —,  and  -~  are  equal  to  the  values  of  -^  at  A,  B, 
dz   dz  dz  dx 

and  F,  respectively;  therefore,  from  equation  (i)  of  Art.  130, 

a'  =  H^b'  =  H^,f'  =  H^.- 
dz  dz  dz 


BEAM   SUSTAINING   MOVING    LOADS. 

hence  ^Jja'+ljb'-f't 

or,  comparing  with  equation  (4), 

^=g>-f'  =  FiG>.  ...........  (6) 

132.  Condition  for  Maximum  or  Minimum  Bending  Moment  at 
Any  Section.  —  As  the  beam  moves  relatively  to  the  loads,  FG 
(proportional  to  M)  in  general  varies.  As  it  passes  through  a 

maximum  or  a  minimum  value,  ——  must  equal  zero  ;  but,  from 

dz 

the  relation  just  proved,  this  makes  F'G'  =  o.     That  is, 

When  the  bending  moment  at  any  given  section  of  the  beam  is 
a  maximum  or  a  minhmim,  the  Hue  A'  B'  intersects  the  load 
polygon  at  a  point  directly  opposite  the  section  considered. 

This  condition  may  be  stated  in  algebraic  form  as  follows  : 
Let   W^  denote  the  total  load  between  Al  and  Fv    W2  the 

load  between  Fl  and  Blt  and  W  the  total  load  on  the  beam. 

Evidently, 

W^f'-a';    W^b'-f-    W=b'-a'. 

If  the  points  F'  and  G'  coincide,/'  —  a'  and  b'  —  f  are  propor- 
tional to  A'  G'  and  G'B'  ,  that  is  to  /j  and  /2,  and  therefore, 


That  is,  when  the  bending  moment  at  any  section  of  the  beam 
is  a  maximum  or  a  minimum,  the  average  load  per  unit  length 
on  each  segment  is  equal  to  the  average  load  per  unit  length  on 
the  whole  span.* 

*  It  is  to  be  noted  that  if,  at  any  point,  there  is  a  load  which  is  concentrated  in  the  strict 

mathematical  sense  (i.e.,  a  finite  load  applied  at  a  mathematical  point),  the  value  of  -~- 

dx 
suffers  a  sudden  change  of  value  at  the  point  of  application  of  that  load,  and  the  equation 


does  not  hold  at  that  point.    This  consideration  does  not,  however,  invalidate  the  above 
conclusions.     For  (a)  a  concentrated  load  in  the  strict  mathematical  sense  does  not  occur 


112  GRAPHIC   STATICS. 

133.   Discrimination    between    Maxima   and   Minima.  —  The 

magnitude  of  the  bending  moment  is 


it  is  obvious  from  the  form  of  the  funicular  polygon  that  g  is 
always  greater  than  /. 

If  g1  —  f  is  positive,  g—  f  (and  therefore  M)  increases  as  z  in- 
creases (i.e.,  as  the  beam  is  displaced  in  the  positive  direction). 
If  g'  —f  is  negative,  M  decreases  as  s  increases.  Therefore, 
as  M  passes  through  a  maximum  value,  the  sign  of  g'  —f 
changes  from  plus  to  minus  ;  and  as  M  passes  through  a  mini- 
mum value,  g'  —f  changes  from  minus  to  plus  (the  displace- 
ment of  the  beam  being  in  the  positive  direction,  that  is,  toward 
the  right  as  the  figure  is  drawn  on  PI.  IV). 

If  the  load  series  consists  of  concentrated  loads,  g'  —f  will  in 
general  pass  through  zero  only  when  a  load  is  at  one  of  the 
three  points  Alt  £v  F^.  By  applying  the  test  just  given,  it  may 
be  seen  that  a  maximum  value  can  result  only  when  a  load 
passes  Fv 

In  applying  the  criterion  for  a  position  of  maximum  or  mini- 
mum value  of  the  bending  moment,  and  in  distinguishing  posi- 
tions causing  maximum  stress  from  those  causing  minimum 
values,  it  is  not  necessary  to  draw  the  line  A'B'  for  every  trial 
position  of  the  beam.  A  movable  strip  of  paper  representing 
the  beam  may  be  laid  in  any  desired  position,  the  points  A'  and  B' 
approximately  located  by  inspection,  and  a  thread  stretched 
between  them.  If  the  thread  crosses  the  load  polygon  at  a 
point  directly  opposite  the  section  in  question,  the  condition  for 

in  practice,  and  (b)  the  ideal  case  of  a  finite  load  applied  at  a  mathematical  point  may  be 
treated  as  the  limit  of  a  case  of  distributed  loading,  so  that  even  in  that  case  the  conclu- 
sions above  derived  hold,  with  slight  change  in  the  form  of  statement. 

The  series  of  concentrated  loads  actually  coming  upon  a  bridge  in  practice  gives  a  load 
polygon  which,  while  approaching  the  form  shown  on  PI.  IV7,  differs  from  it  in  that  the 
portions  under  the  loads  are  not  strictly  vertical  lines,  but  are  lines  of  variable  (very  steep) 
inclination. 

In  applying  the  condition  just  deduced,  a  load  at  either  of  the  three  points  A\,  BI,  F\t 
is  to  be  regarded  as  distributed  in  an  arbitrary  way  along  the  small  element  of  horizontal 
distance  near  the  line  used  as  its  line  of  action. 


BEAM   SUSTAINING   MOVING   LOADS.  113 

maximum  or  minimum  is  fulfilled.  Only  positions  which  bring 
a  load  at  the  section  need  be  tried.  The  accurate  location  of 
A'  and  B'  is  usually  unnecessary.* 

At  (B\  PI.  IV,  is  shown  the  beam  AlBl  in  such  a  position 
that  F'G'=o.  It  is  seen  that  this  position  corresponds  to  a 
maximum  value  of  the  bending  moment  at  F1 ;  for  a  displace- 
ment of  the  beam  toward  the  right  (positive)  makes  g1  — f 
negative,  while  a  displacement  toward  the  left  makes  g'  —f 
positive. 

134.  Determination  of  Bending  Moment.  —  Having  determined 
the  position  of  the  load  series  causing  greatest  bending  moment 
at  a  given  section,  the  value  of  the  bending  moment  may  be 
computed  by  the  method  already  explained  in  the  case  of  fixed 
loads.  At  (M),  PI.  Ill,  is  represented  the  position  of  the  beam 
A-iB-fr  which  causes  the  greatest  bending  moment  at  the  section 
Fv  distant  16  ft.  from  Alt  the  span  being  64  ft.  That  the  con- 
dition for  a  maximum  is  satisfied  in  this  position  may  easily  be 
verified.  The  total  load  on  the  beam  is  112000  Ibs.,  one-fourth 
of  which  is  28000  Ibs.  By  regarding  5000  Ibs.  of  the  load  3-4 
as  belonging  to  the  segment  A-^F^  the  load  on  this  segment  is 
28000  Ibs.  It  is  seen  also  that  the  beam  is  fully  loaded,  and 
that  the  heaviest  loads  are  near  the  section.  Moreover,  the 
condition  for  maximum  is  not  satisfied  for  any  other  position 
near  the  one  chosen. 

The  closing  string  of  the  funicular  polygon  for  all  external 
forces  acting  on  the  beam  in  the  position  (M)  is  AB.  The 
bending  moment  at  F^  is  the  product  of  the  ordinate  FG  by  the 
pole  distance. 

At  (B\  PI.  Ill,  is  shown  the  diagram  of  maximum  bending 
moments  for  all  sections  of  the  beam.  From  AlBl  are  laid  off 
downward  ordinates  proportional  to  the  maximum  bending 
moments.  The  length  FG  just  determined  is  laid  off  from  Fl; 
this  must  be  multiplied  by  the  pole  distance  (100000  Ibs.  in  the 

*  See  paper  by  Professor  Henry  T.  Eddy,  Trans.  Am.  Soc.  C.  E.,  Vol.  XXII, 


II4  GRAPHIC   STATICS. 

diagram).  The  complete  curve  is  drawn,  but  the  construction 
for  determining  values  of  the  maximum  bending  moment  is 
shown  only  for  the  section  Fv 

135.  Moment  Curve  for  Fixed  Loads.  — The  greatest  bending 
moment  due  to  moving  loads  must  be  combined  with  the  bend- 
ing moment  due  to  fixed  loads.     If  the  fixed  load  is  uniformly 
distributed,  as  already  assumed  in  the  computation  of  shear, 
it  may  be  divided  into  parts,  each  assumed  concentrated  at 
its  center  of  gravity,  and  a  funicular  polygon  drawn,  using  the 
same  pole  distance  employed  in  the  force  diagram  for  moving 
loads.     The  ordinates  of  this  funicular  polygon  may  be  laid  off 
upward  from  the  line  A-^B^  and  their  ends  joined  to  form  a 
continuous  curve.     The  total  ordinate  from  this  curve  to  that 
already  drawn  for  moving  loads  represents  the  true  greatest 
bending  moment  at   the  corresponding  section  of   the   beam. 
The  curve  is  shown  at  (B),  PL   III,  but  the  construction  is 
omitted,  since  it  involves  no  new  principle. 

It  is  to  be  remembered  that  the  bending  moment  found  for 
any  section  is  a  possible  value  for  the  other  section  equally 
distant  from  the  center  of  the  beam,  since  the  train  may  be 
headed  in  the  opposite  direction,  and  the  same  construction 
made,  viewing  the  beam  from  the  other  side.  The  same  state- 
ment holds  as  to  shears. 

136.  Design  of  Beam  sustaining  Moving  Loads.  —  In  designing 
a  beam  to  sustain  moving  loads,  the  greatest  shear  and  bending 
moment  that  can  come  upon  it  for  any  position  of  the  loads  must 
be  known  for  every  section.      The  methods  above  given  are 
sufficient  for  the  determination  of  these  quantities;    and   the 
problem  of  designing  the  beam  will  not  be  here  further  discussed. 


CHAPTER    VII.      TRUSSES    SUSTAINING    MOVING 
LOADS. 

§   I.    Bridge  Loads. 

1 37.  General  Statement.  — The  most  important  class  of  trusses 
sustaining  moving  loads  is  that  of  bridge  trusses.     The  two 
main  classes  of  bridges  are  highway  bridges  and  railway  bridges. 
The  forms  of  trusses  most  commonly  used  differ  for  the  two 
classes,  as  do  also  the  amount  and  distribution  of  loads. 

Before  the  design  can  be  correctly  made,  the  weights  of  the 
trusses  themselves  must  be  known.  Since  these  weights  de- 
pend upon  the  dimensions  of  the  truss  members,  they  cannot 
be  known  with  certainty  until  the  design  is  completed.  The 
remarks  made  in  Art.  82  regarding  the  design  of  roof  trusses 
are  here  applicable. 

In  the  following  articles  we  shall  give  data  available  for 
preliminary  estimates  of  truss  weights. 

Railway  bridges  of  short  span  are  frequently  supported  by 
rolled  or  built  beams.  When  the  span  is  longer  than  100  ft. 
trusses  should  be  used.  (Cooper's  "  Specifications  for  Iron  and 
Steel  Railroad  Bridges.") 

138.  Loads  on  Highway   Bridges.  —  (i)  Permanent  load. — 
The  permanent  load  sustained  by  a  highway  bridge  truss   in- 
cludes the  weight  of  the  truss  itself,  of  the  lateral  or  "sway" 
bracing,  of  the  floor  and  the  beams  and  stringers  supporting  it. 
These  weights  are  all  subject  to  much  variation,  but,  for  pur- 
poses of  preliminary  design,  the  following  formula,  taken  from 
Merriman's  "  Roofs  and  Bridges,"  may  be  used. 

"5 


H6  GRAPHIC   STATICS. 

Let  w  =  weight  of  bridge  in  pounds  per  linear  foot;  /=span 
in  feet ;  b  =  width  in  feet.  Then 

w=  140  + 12  b  +  o.2  £/— 0.4  /. 

(2)  Snoiv  load.  —  The  weight  of  snow  may  be  taken  as  in 
case  of  roof  trusses   (Art.   84).     The  values  there  given    are 
probably  in  excess  of  those  ordinarily  employed  in  practice. 

(3)  Wind  load.  —  The  pressure  of  wind  striking  the  bridge 
laterally  is  resisted  by  the  chord  members  together  with  the 
lateral  bracing.     These  constitute  horizontal  trusses,  in  which 
the  stresses  are  to  be  found  in  the  same  way  as  for  the  main 
trusses  of  the  bridge.     As  the  determination  of  wind  stresses 
requires  the  use  of  no  special  methods  or  principles,  they  will 
not   be   here   considered.     The   student  is  referred  to   Burr's 
"  Stresses   in    Roofs   and    Bridges,"    Merriman's    "  Roofs    and 
Bridges,"  and  other  available  works  for  a  complete  discussion 
of  wind  pressure  and  its  effects  on  bridge  trusses. 

(4)  Moving-  load.  —  The  most  dangerous  moving  load  for  a 
highway  bridge  is  usually  a  crowd  of  people   or   a  drove  of 
animals.     This  is  commonly  taken  as  a  uniformly  distributed 
load,  which  may  cover  the  whole  bridge  or  any  portion  of  it. 
Its  value  is  variously  taken  at  from  60  Ibs.  to  100  Ibs.  per  square 
foot  of  area  of  floor,  depending  upon  the  span  and  upon  local 
conditions. 

It  may  be  that  in  certain  cases  the  greatest  stresses  will 
result  from  the  passage  of  heavy  pieces  of  machinery  over  the 
bridge,  as,  for  example,  a  steam  road  roller.  This  should  of 
course  be  considered  in  the  design. 

For  a  complete  discussion  of  loads  on  highway  bridges,  the 
student  is  referred  to  Waddell's  "  Highway  Bridges." 

139.  Loads  on  Railway  Bridges. — (i)  Permanent  load. — 
The  permanent  load  on  a  railway  bridge  includes  (a)  the  weight 
of  the  track  system,  which  is  known  or  may  be  determined  at 
the  outset ;  (b)  the  weight  of  longitudinal  stringers  and  cross- 


BRIDGE   LOADS.  117 

beams,  which  can  be  determined  before  the  trusses  are  designed  ; 
and  (c)  the  weights  of  trusses.  The  weight  of  the  track  system 
may  be  taken  at  400  Ibs.  per  linear  foot  for  a  single  track. 
(See  Burr's  "  Stresses  in  Bridge  and  Roof  Trusses  "  ;  Cooper's 
"  Specifications  for  Iron  and  Steel  Railroad  Bridges  "  ;  Merri- 
man's  "  Roofs  and  Bridges.")  The  total  weight  of  track  system 
and  supporting  beams  and  stringers  varies  from  450  Ibs.  to 
600  Ibs.  per  linear  foot.  (Merriman.)  For  spans  less  than  100 
feet,  Merriman  gives  the  following  formulas,  in  which  w  is  the 
total  dead  load  of  the  bridge  in  pounds  per  linear  foot,  and  /  is 
the  span  in  feet : 

For  single  track,   w=   560+    5.6 /. 

For  double  track,  w=  1070+  10.7  /. 
See  also  Art.  8  of  Burr's  work  above  cited. 

(2)  Snow  and  wind.  —  Railway  bridges   usually  offer  little 
opportunity  for  the  accumulation  of  snow.     Wind  pressure  is, 
however,  an  important  factor.     Besides  the  pressure  upon  the 
bridge  itself,  the  pressure  upon  trains  crossing  the  bridge  must 
be  considered.     The  latter  is  a  moving  load  and  may  be  dealt 
with  in  the  same  way  as  other  moving  loads.     Its  amount  may 
be  computed  from  the  area  of  the  exposed  surface  of  the  train 
and  the  known   (or  assumed)   greatest  pressure  due  to  wind 
striking  a  vertical  surface  (Art.  85). 

For  further  discussion  of  wind  pressure,  the  student  is  referred 
to  the  works  already  cited.  The  graphic  methods  of  deter- 
mining stresses  due  to  wind  will  be  evident  when  the  methods 
for  vertical  loads  given  in  the  following  articles  are  understood. 

(3)  Moving  loads.  —  The  moving  load  to  be  supported  by  a 
bridge  consists  of  the  weights  of  trains.     Such  a  load  is  applied 
to  the  track  at  a  series  of  points,  namely,  the  points  of  contact 
of  the  wheels.     But  the  load  is  applied  to  the  trusses  only  at 
the  points  at  which  the  floor  beams  are  supported.     Hence  the 
actual  distribution  of  loads  upon  the  truss  is  somewhat  com- 
plex.    It  was   formerly  common   to   substitute  for  the  actual 


H8  GRAPHIC   STATICS. 

load  a  uniformly  distributed  load,  thus  simplifying  the  problem 
of  'determining  stresses.  It  is  now  more  usual  to  consider  the 
actual  distribution  of  loads  for  some  standard  type  of  locomo- 
tive used  by  the  railroad  concerned,  or  specified  by  its  engi- 
neers. For  examples  of  such  distributions  the  student  is 
referred  to  Cooper's  "Specifications"  already  cited;  also  to 
PI.  Ill,  and  to  the  following  portions  of  this  chapter. 

140.  Through  and  Deck  Bridges.  —  A  bridge  is  called  through 
or  deck  according  as  the  floor  system  is  supported  at  points  of 
the  lower  or  of  the  upper  chord.     In  the  former  case,  if  the 
trusses  are  too  low  to  require  lateral  bracing  above,  they  are 
called  pony  trusses. 

The  weight  of  the  truss  itself  is  to  be  divided  between  the 
upper  and  lower  joints.  But  the  weight  of  the  floor  system 
and  of  the  supporting  beams  and  stringers  comes  wholly  at  the 
lower  joints  of  a  through  bridge,  or  at  the  upper  joints  of  a 
deck  bridge.  The  moving  load  is,  of  course,  applied  at  the 
same  joints  at  which  the  floor  system  is  supported. 

If  the  floor  system  is  supported  directly  upon  the  upper 
chord,  as  is  sometimes  the  case,  the  moving  load  and  part  of 
the  dead  load  produce  bending  in  the  chord  members ;  the 
design  is  otherwise  unaffected  by  this  construction. 

§  2.    Truss  regarded  as  a  Beam. 

141.  Classification  of  Trusses.  —  Since  a  bridge  truss  acts  as 
a  practically  rigid  body  resting  on  supports  at  the  ends  or  other 
points  and  sustaining  vertical  loads,  it  may  be  regarded  as  a 
beam,  and  trusses  may  be  classified  in  the  same  way  as  beams 
(Art.   114).     The  only  class  to  be  here  considered  is  that  of 
simple  trusses,  or  such  as  may  be  regarded  as  rigid  bodies  for 
the  purpose  of  determining  the  reactions. 

Cantilever  trusses  and  continuous  trusses  are  defined  like  the 
corresponding  classes  of  beams  (Art.  114).  The  most  impor- 
tant case  is  that  of  a  truss  simply  supported  at  the  ends. 


TRUSS   REGARDED   AS   A   BEAM. 

142.  External  Shear  for  a  Truss.  —  If  a  truss  be  re^ 
a  beam,  the  external  shear,  resisting  shear,  and 

ing  stress  at  any  section  may  be  denned  just  as  in  Art.  iT$? 
In  some  forms  of  truss  a  knowledge  of  the  external  shear  at 
any  section  makes  it  possible  to  compute  readily  the  stresses  in 
certain  truss  members.  Thus,  in  the  portion  of  a  truss  repre- 
sented in  Fig.  47,  let  the  section 
MN  cut  three  members,  of  which 
two  are  horizontal.  (The  member 
x'y'  is  disregarded.)  Since  each 
member  can  exert  forces  only  in 
the  direction  of  its  length,  the  ex- 
ternal shear  in  the  section  MN 
must  be  wholly  resisted  by  the  di- 
agonal xy ;  and  the  internal  force 
in  xy  must  be  such  that  its  resolved 

part  in  the  vertical  direction  is  equal  in  magnitude  to  the  exter- 
nal shear  in  the  section.  Let  the  external  shear  V  be  repre- 
sented by  YZ  (Fig.  47) ;  draw  YX  parallel  to  yx  and  ZX 
horizontal;  then  JfFwill  represent  the  stress  in  xy.  If  V  is 
positive  (Art.  115),  the  stress  in  xy  is  a  tension.  If  Fis  nega- 
tive, the  stress  is  a  compression.  If  the  member  xy  were 
replaced  by  one  sloping  the  other  way  from  the  vertical,  these 
statements  as  to  kind  of  stress  would  be  reversed. 

If  no  two  of  the  three  members  cut  by  any  section  are  par- 
allel, the  stresses  cannot  be  computed  so  simply,  since  all  may 
contribute  components  of  force  to  resist  the  external  shear. 

143.  Bending   Moment  for   a   Truss.  —  In   many  cases  the 
stresses  in  the  truss  members  can  be  found  from  the  values  of 
the  bending  moment  at  different  sections. 

Thus,  in  Fig.  49,  let  a  section  MN  be  taken  cutting  three 
members  as  shown,  and  let  the  origin  of  moments  be  taken  at 
the  point  of  intersection  of  two  of  them  (as  bx  and  xy)  ;  then 
since  the  moments  of  the  internal  forces  in  these  two  will  be 


120  GRAPHIC    STATICS. 

zero,  the  resisting  moment  is  equal  to  the  moment  of  the  inter- 
nal force  in  the  third  member  ay.  The  arm  of  this  latter  force 
is  the  perpendicular  distance  of  its  action  line  from  the  origin. 
Call  it  h,  and  let  /  represent  the  force  itself,  and  M  the  bending 
moment.  Then,  numerically, 


If  M  is  positive  (in  accordance  with  the  convention  of  Art. 
47),  /  must  act  from  right  to  left,  hence  the  stress  in  ay  is  a 
compression.  If  M  is  negative,  the  stress  is  a  tension.  (It 
must  be  remembered  that  the  forces  whose  moments  make  up 
the  bending  moment  M  act  upon  the  portion  of  the  truss  to 
the  left  of  the  section.) 

§  3.     Truss  sustaining  Any  Series  of  Moving  Loads. 

144.  General  Method  of  Determining  Maximum  Stresses  in 
Truss  Members.  —  When  a  truss  carries  a  definite  series  of  mov- 
ing loads,  the  problem  of  determining  the  maximum  stress  in 
any  member  involves  (i)the  determination  of  the  proper  posi- 
tion of  the  loads  and  (2)  the  actual  computation  of  the  stress  for 
a  known  position  of  the  loads.  The  second  part  of  this  problem 
involves  only  the  principles  already  explained  in  the  discussion 
of  trusses  carrying  fixed  loads.  The  first  part  requires  special 
consideration. 

Although  the  treatment  of  trusses  with  parallel  chords  is  in 
some  respects  simpler  than  that  of  the  case  of  non-parallel 
chords,  it  is  advantageous  to  consider  the  problem  in  as  general 
a  manner  as  possible.  The  following  discussion  will  therefore 
refer  to  the  general  case  of  a  truss  of  any  form,  subject  only  to 
the  restriction  that  there  are  no  redundant  members,  so  that  the 
truss  may  be  completely  divided  by  cutting  three  members,  of 
which  any  chosen  member  may  be  one.*  No  restriction  will  be 

*  A  case  not  included  in  this  general  description,  while  still  offering  a  determinate 
problem,  is  that  of  a  truss  with  subordinate  bracing.  This  important  case  will  be  consid- 
ered in  a  following  discussion. 


TRUSS  SUSTAINING  ANY  SERIES  OF  MOVING  LOADS.     121 

made  as  to  the  distribution  of  the  loads.  Trusses  with  parallel 
chords,  and  particular  distributions  of  loading,  may  be  treated 
as  special  cases  of  the  general  problem.* 

145.  Computation  of  Stress  when  Position  of  Loads  is 
Known.  —  The  general  "  method  of  sections  "  (Art.  69)  may  be 
applied  to  determine  the  stress  in  any  member  when  the  posi- 
tion of  the  load  series  is  known. 

Thus,  referring  to  PL  V,  let  1-2,  2-3,  . . .  represent  the  lines 
of  action  of  the  successive  loads,  1-2-3-  •  •  •  the  force  polygon 
for  the  loads,  and  let  a  funicular  polygon  be  drawn  as  shown. 
The  truss  is  represented  in  two  positions  marked  (A)  and  (B} 
respectively.  In  connection  with  (A)  is  shown  the  construction 
for  determining  the  stress  in  the  member  C-JD^,  and  in  connec- 
tion with  (B}  the  construction  for  determining  the  stress  in  C^H. 
Referring  to  the  former  case,  let  a  section  be  taken  through  the 
three  members  C-J)^  C-^F^  H\F<L  5  then  the  internal  forces  in 
the  three  members  cut  are  in  equilibrium  with  the  external 
forces  acting  on  the  portion  of  the  truss  on  either  side  (as  the 
left)  of  the  section.  These  forces  are  the  reaction  at  A^  and  all 
the  loads  from  Al  to  C^.  Applying  the  principle  of  moments,  the 
origin  may  be  taken  at  F2,  the  point  of  intersection  of  the  lines 
C±FZ  and  H^F^  thus  eliminating  the  forces  acting  in  these  lines. 

The  moment  of  any  load  is  equal  to  the  sum  of  the  moments 
of  any  components  which  may  replace  it.  Hence  for  any  load 
between  Al  and  Cl  (though  actually  coming  upon  the  truss  at 
the  panel  joints)  the  moment  may  be  computed  as  if  it  were 
applied  to  the  truss  at  a  point  in  the  vertical  line  through  its 
actual  point  of  application  on  the  floor  or  track.  But  in  the 
case  of  a  load  on  the  panel  -C-J)-^  only  one  component  comes 
upon  the  portion  of  the  truss  to  which  the  moment  equation  is 
to  be  applied  ;  it  is  therefore  necessary  to  determine  the  portion 
of  every  such  load  which  is  actually  carried  at  Cv  If  loads 
between  C±  and  Dl  are  carried  wholly  at  the  two  points  Cl  and 

*  The  case  of  parallel  chords,  with  graphic  methods  of  treatment,  receives  very  full  dis- 
cussion in  a  paper  by  Professor  Henry  T.  Eddy,  Trans.  Am.  Soc.  C.  E.,  Vol.  XXII  (1890). 


122  GRAPHIC    STATICS. 

D±t  the  portion  carried  at  each  of  these  points  is  easily  deter- 
mined, as  will  be  shown.* 

From  A1  and  Bl  draw  vertical  lines  intersecting  the  funicular 
polygon  at  A  and  B  respectively  ;  the  line  AB  is  the  closing 
line  of  the  funicular  polygon  for  the  external  forces  acting  upon 
the  truss  when  the  load  series  has  the  assumed  position.  On 
the  panel  ClDl  is  the  load  6' 7' '.  Treating  C1D1  as  a  simple 
beam,  the  components  replacing  this  load  at  C1  and  D1  are  found 
(Art.  45)  as  follows :  Draw  vertical  lines  through  Cl  and  Dl 
intersecting  the  strings  06'  and  of  in  C  and  D ;  from  the  pole  O 
draw  the  ray  OK  parallel  to  CD ;  the  required  components  of 
6'7'  are  represented  in  the  force  polygon  by  6' K  and  Kf.  The 
same  construction  might  be  made  for  every  panel,  and  it  is  seen 
that  the  series  of  lines  such  as  CD  would  form  the  funicular 
polygon  for  the  loads  as  actually  applied  to  the  truss  at  the 
several  joints.  But  the  construction  is  here  needed  only  for 
the  panel  C-J)lt  since  a  load  on  any  other  panel  comes  wholly 
upon  one  of  the  two  portions  into  which  the  truss  is  divided  by 
the  section  taken. 

The  sum  of  the  moments  of  the  external  forces  acting  upon 
the  left  portion  of  the  truss  may  be  found  graphically  by  the 
method  of  Art.  56.  That  is,  through  the  origin  of  moments  a 
line  is  drawn  parallel  to  the  resultant  of  the  forces  (a  vertical 
line  in  this  case) ;  its  intercept  FG,  between  the  strings  AB  and 
CD,  multiplied  by  the  pole  distance,  gives  the  required  moment. 
The  origin  being  taken  at  F2,  this  moment  is  equal  to  the 
moment  of  the  internal  force  in  the  member  C-J)^ ;  hence  that 
force  may  be  found  by  dividing  the  moment  by  the  perpendicu- 
lar distance  from  Fz  to  C-JDV 

Referring  next  to  {B)  and  considering  the  member  C-^H,  the 
stress  in  this  member  may  be  found  in  a  similar  manner.  The 

*  It  is  to  be  noticed  that  the  actual  distribution  of  any  load  among  the  different  joints  is 
not  determinate  by  simple  methods,  if  the  longitudinal  beams  supporting  the  floor  system 
are  supported  at  more  than  two  points.  It  will  here  be  assumed  that  the  portion  of  such  a 
beam  between  two  supports  acts  as  a  simple  beam.  The  results  of  this  assumption  are 
probably  as  reliable  as  could  be  obtained  by  a  more  elaborate  discussion. 


U.  OF  C. 

TRUSS  SUSTAINING  ANY  SERIES  OF  MOVING 


points  A,  B,  C,  D,  F,  G  have  the  same  meanings  in  this  con- 
struction as  in  the  preceding,  but  AB  and  CD  must  be  produced 
in  order  to  find  the  intercept  FG,  because  the  origin  of  moments 
is  at  Flt  beyond  the  end  of  the  span.  The  moment  of  the 
internal  force  in  C^H  is  equal  to  FG  multiplied  by  the  pole  dis- 
tance ;  dividing  this  moment  by  the  perpendicular  distance 
from  Fl  to  C^H  gives  the  magnitude  of  the  required  force. 

The  kind  of  stress  in  any  member  may  be  determined  by 
inspection  of  the  conditions,  as  in  Art.  in. 

If  the  stress  in  a  member  is  determined  by  the  foregoing 
method,  the  steps  of  the  process  are  the  same,  whatever  mem- 
ber be  under  consideration.  These  steps  are  as  follows : 

(1)  A  section  is  taken  completely  dividing  the  truss  and  cut- 
ting only  three  members,  one  of  which  is  the  member  whose 
stress  is  to  be  determined,  and  one  a  member  of  the  chord  car- 
rying moving  loads. 

(2)  Taking  origin  at  the  intersection  of  those  two  of  the 
three  members  cut  whose  stresses  are   not   in    consideration, 
equate  the  sum  of  the  moments  of  the  external  forces  acting 
upon  one  portion  of  the  truss  to  the  moment  of  the  internal 
force  in  the  member  considered. 

146.  Definitions  and  Notation.  —  The  following  definitions 
and  notation  will  conduce  to  clearness  and  generality  in  the  ensu- 
ing discussion. 

Let  the  origin  of  moments,  assumed  in  determining  the  stress 
in  any  member  by  the  foregoing  general  method,  be  called  the 
moment-center  for  that  member. 

Let  the  stress-moment  for  any  member  be  defined  as  the  mo- 
ment of  the  stress  in  that  member  with  respect  to  the  moment- 
center  just  described.  Let  the  sign  of  the  stress-moment  be 
specified  in  the  following  manner  : 

Rotation  being  called  positive  or  negative  in  accordance  with 
the  convention  adopted  in  Art.  47,  let  the  stress-moment  be 
called  positive  when  it  resists  a  tendency  of  the  left  portion  of 


124 


GRAPHIC   STATICS. 


the  truss  to  rotate  negatively  (and  of  the  right  portion  to  rotate 
positively). 

In  -all  cases  let  the  left  and  right  ends  of  the  span  be  marked 
Al  and  Bv  and  let  C±  and  D1  designate  the  left  and  right  ends 
of  the  panel  cut  by  the  assumed  section.  Let  the  vertical  line 
through  the  moment-center  cut  A1ff1  (produced  if  necessary)  in 
a  point  marked  Fv 

Distances  from  left  to  right  being  positive,  and  from  right  to 
left  negative,  let  the  following  notation  be  used  : 


C-fi  =  «lt  F1Dl  =  n2, 
The  sign  of  each  of  the  four  quantities  /1(  /2,  »lf  «2  is  to  conform 
in  all  cases  to  the  direction,  as  indicated  by  the  order  of  the  letters 
used  in  defining  it.  Thus  /1  =  A1F1  is  positive  or  negative, 
according  as  F1  is  to  the  right  or  to  the  left  of  Av 

147.  Effect  of  a  Single  Load.  —  Influence  line.  —  A  general 
idea  of  the  position  of  a  load  series  which  will  cause  the  greatest 
stress  in  any  member  of  a  truss  may  be  obtained  by  considering 
the  effect  of  a  single  load  placed  in  any  position.  The  details 
of  the  discussion  will  vary  with  the  form  of  the  truss,  but  the 
method  of  reasoning  may  be  sufficiently  illustrated  by  reference 
to  the  form  shown  at  (A)  or  (B),  PL  V. 

The  relative  effect  of  loads  in  different  positions  in  producing 
stress  in  a  given  member  may  be  clearly  represented  by  means 
of  an  "  influence  diagram,"  analogous  to  the  diagrams  used  in 
the  discussion  of  shear  and  bending  moment  in  a  beam  (Arts. 
126  and  129). 

From  a  given  base  line  let  ordinates  be  erected  such  that  the 
ordinate  at  any  point  represents  (to  a  convenient  scale)  the  stress 
due  to  a  unit  load  at  the  corresponding  point  of  the  span.  The 
line  joining  the  extremities  of  these  ordinates  is  the  influence 
line  *  for  the  member  in  question. 

*  For  a  discussion  of  influence  lines,  see  paper  by  Professor  George  F.  Swain,  Trans. 
Am.  Soc.  C.  E.,  Vol.  XVII,  p.  21  (1887). 


TRUSS  SUSTAINING  ANY  SERIES  OF  MOVING  LOADS.      I25 

It  will  be  shown  that  there  is  a  simple  method,  of  complete 
generality,  by  which  the  influence  line  may  be  drawn  for  any 
member  whatever.  In  order  to  explain  this  method  clearly,  it 
is  necessary  to  adopt  a  convention  as  to  signs. 

In  laying  off  ordinates  of  the  influence  line,  let  positive  values 
be  drawn  upward  and  negative  values  downward  from  the  refer- 
ence line.  Let  the  sign  of  the  ordinate  agree  in  all  cases  with 
tJiat  of  the  stress-moment  (Art.  146).  With  this  convention,  a 
positive  ordinate  may  correspond  to  either  a  compressive  or  a 
tensile  stress,  depending  upon  the  relation  of  the  moment-center 
to  the  member  under  consideration.  F"rom  the  sign  of  the  stress- 
moment  the  kind  of  stress  may  always  be  readily  determined  by 
inspection  of  the  truss  diagram. 

In  Fig.  48  (PL  VIII)  are  shown  the  forms  assumed  by  the 
influence  line  for  three  typical  members  of  the  truss  shown; 
(A)  represents  the  case  of  the  chord  member  C-J)-^  (B)  that  of 
the  chord  member  C^D^,  and  (Q  that  of  the  web  member 
C^ Dv  The  method  of  constructing  these  diagrams  will  now 
be  explained. 

Member  of  loaded  chord.  —  Consider  first  the  member  C-J)^ 
The  moment-center  is  6\',  and  F1  falls  between  C±  and  D^,  the 
corresponding  position  of  Fz  being  shown  at  (A).  Here  llt  /2, 
;/lf  «2  are  all  positive  (Art.  146).  A  load  P  between  D1  and  B-^ 
distant  x  from  B^  causes  a  reaction  —  at  Av  and  a  stress- 
moment  equal  to  — ^-.  This  varies  directly  as  x,  being  o  at  JB1 
and  — 1(  2~;?2)  at  /^  Similarly,  the  stress-moment  due  to  a 
load  P  between  Al  and  Cv  distant  x  from  AI}  is  — ^—,  which 
varies  from  o  at  A1  to  2^  i~*i)  at  Cv  If  the  moment-arm 

of  the  required  stress  is  /,  the  ordinates  of  the  influence  line  at 
C2  and  Z?2  are 

Ct,C •>•=•  — — —, — —  and 


and  the  straight  lines  AZCS  and  D^BZ  are  the  portions  of  the 


126  GRAPHIC   STATICS. 

influence  line  for  the  corresponding  portions  of  the  span.     A 
load  P  between  C±  and  Dv  distant  x  from  Clt  comes  upon  the 

truss  partly  at  £\  and  partly  at  D1 ;  the  former  part  is  — *— — —> 


Px 


n 


the  latter  —  .     The  total  stress-moment  due  to  such  a  load  is 
therefore 


putting  jP=i  and  dividing  by  /,  this  expression  gives  the  ordi- 
nate  of  the  influence  line  for  any  point  between  C\  and  D±.  The 
influence  line  for  this  portion  of  the  span  is  therefore  the  straight 
line  C3D3.  It  will  be  noticed  that,  if  the  straight  line  B^D^  be 
continued,  its  ordinate  at  Fz  will  be  F^f^  =  ~^\  if  AZC3  be  con- 
tinued, its  ordinate  at  F2  will  have  the  same  value. 

Member  of  chord  not  carrying  moving  loads.  —  Next  consider 
the  member  CJDJ.  The  moment-center  is  Z\,  F1  falls  at  D^ 
and  F2  coincides  with  Z>2  as  shown  at  (/?),  Fig.  48.  If  the 
reasoning  used  in  the  preceding  case  be  repeated,  it  is  seen  that 
the  values  there  given  for  the  ordinates  of  the  influence  line 
apply  also  to  this  case.  Since  in  the  present  case  n2=o,  the 
value  of  the  ordinate  at  Z>2  reduces  to  -£-1.  Further,  it  is  seen 
that  A2C3DS  is  a  straight  line. 

Web  member.  —  For  the  member  C1'D1  the  moment-center  is 
at  the  intersection  of  Cj_DJ  and  C-^D^  ;  F±  falls  to  the  left  of  Av 
the  corresponding  position  of  F2  being  shown  at  (C),  Fig.  48. 
Here  it  will  be  noticed  that  /x  and  n^  are  negative.  If,  however, 
the  method  used  in  the  preceding  cases  is  again  applied,  it  is 
found  that  the  values  above  given  for  the  stress-moment  due  to 
loads  on  D-^B-^  and  A1Cl  are  correct  for  the  present  case  both  in 
magnitude  and  in  sign.  Thus,  for  a  load  P  on  D^B^  distant 
x  from  BI,  the  reaction  at  Al  is  —  ,  and  the  stress-moment  is 


—-.     This  stress-moment  is  negative  (according  to  the  conven- 
tion adopted  in  Art.  146)  ;  but  since  ^  is  negative,  the  expres- 


TRUSS  SUSTAINING  ANY  SERIES  OF  MOVING  LOADS.     127 

sion  —  i^  is  correct  in  magnitude  and  sign.     For  a  load  P  on 

A-iC-i,  distant  x  from  Alt  the  stress-moment  is  positive,  and  is 
given  in  magnitude  and  sign  by  the  same  expression  as  in  the 
preceding  cases,  —f->  The  ordinates  at  Cz  and  Dz  also  have 
the  general  values 


the  latter  expression  being  negative  (because  /j  is  negative  and 
/2  —  «2  positive)  while  the  former  is  positive  (because  /2  and  l\  —  n± 
are  both  positive).  Further,  if  AZC&  and  D%B^  be  prolonged, 
they  have  a  common  ordinate  -L?  at  the  point  F2.  This  ordi- 
nate  is  negative,  agreeing  in  sign  with  /j. 

General  result.  —  It  is  thus  seen  that  the  construction  of  the 
influence  line  may  follow  one  general  rule  in  all  cases.  This 
rule  may  be  stated  as  follows,  designating  points  by  letters,  as 
above  : 

From  the  point  Fz  erect  an  ordinate  of  magnitude  and  sign 

~^.  From  the  extremity  of  this  ordinate  draw  straight  lines  to 
Az  and  B^  the  former  intersecting  the  vertical  through  C2  in  Cs, 
the  latter  intersecting  the  vertical  through  Z>2  in  Z?3.  The  line 
AZC3D3B2  is  the  required  influence  line.* 

148.  Approximate  Position  of  Load-Series  causing  Greatest 
Stress  in  a  Given  Member.  —  Inspection  of  the  influence  dia- 
gram for  a  given  member  enables  the  position  of  loads  causing 
maximum  stress  to  be  estimated  approximately. 

Member  of  chord  carrying  moving'  loads.  —  -  For  the  member 
ClDl  (Fig.  48,  PI.  VIII),  the  influence  diagram  (A)  shows  that 

*  The  same  general  rule  applies  when  F±  falls  between  A±  and  Clt  or  between  Dl  and 
Blt  n1  being  negative  in  one  case,  and  n.2  in  the  other,  while  both  4  and  /2  are  positive  in 
both  cases.  These  cases  occur  in  no  ordinary  form  of  truss,  except  that  with  subordinate 
bracing.  This  form  is  discussed  in  Art.  160,  and  the  influence  diagrams  for  negative  val- 
ues of  «!  and  of  n.2  are  shown  on  PI.  VIII. 


128  GRAPHIC    STATICS. 

all  loads  on  the  truss  produce  the  same  kind  of  stress,  and  that 
the  effect  of  a  given  load  is  greater  the  nearer  it  is  *  to  D^. 
Therefore,  with  any  given  load  series,  it  is  to  be  expected  that 
the  greatest  stress  will  occur  when  (a)  the  truss  is  heavily 
loaded  throughout  the  span,  and  (b)  the  heaviest  loads  are  near 
the  point  Dv 

Chord  not  carrying  moving  loads.  —  In  a  similar  manner,  the 
influence  diagram  (B~)  shows  that  the  greatest  stress  in  C-^D-^  is 
to  be  expected  when  (a)  the  entire  span  is  heavily  loaded,  and 
(#)  the  heaviest  loads  are  near  Dr 

Web  member. —Diagram  (Q,  Fig.  48,  PL  VIII,  shows  that 
for  loads  between  A1  and  El  the  stress-moment  is  positive, 
while  for  loads  between  E1  and  B^  it  is  negative.  Hence  for 
the  maximum  stress  of  the  kind  corresponding  to  a  positive 
stress-moment  (a)  the  portion  A1E1  should  be  heavily  loaded 
while  little  or  no  load  is  between  E1  and  Blt  and  (b}  heavy  loads 
should  be  near  Cv  For  maximum  stress  of  the  kind  corre- 
sponding to  a  negative  stress-moment,  (a}  E1S1  should  be 
heavily  loaded  while  little  or  no  load  is  on  A-^E^,  and  (b}  heavy 
loads  should  be  f  near  Dv 

The  above  principles  serve  as  an  approximate  general  guide, 
but  do  not  enable  us  to  determine  the  position  of  loads  for 
maximum  stress  exactly.  A  more  definite  discussion  of  the 
problem  will  now  be  given. 

149.  Criterion  for  Position  of  Loads  causing  Maximum  or 
Minimum  Stress  in  Any  Member  of  a  Truss.  —  Let  a  funicular 
polygon  and  a  load  polygon  be  drawn  for  the  given  series  of 
loads  (PL  V). 

As  shown  in  Art.  130,  if  y  denotes  any  ordinate  of  the 
funicular  polygon,  and  y'  the  corresponding  ordinate  of  the 

*  It  may,  of  course,  happen  that  C2Q  is  greater  than  D%DS.  In  such  case,  the  effect 
of  a  load  is  greater  the  nearer  it  is  to  Q. 

t  The  nature  of  the  load-series  may  be  such  that  a  compromise  must  be  made  between 
requirements  (a)  and  (6). 


TRUSS  SUSTAINING  ANY  SERIES  OF  MOVING  LOADS.      129 

load    polygon    (the    axes    being    LX,    LY,    and   L'X',   L'Y', 
respectively), 


dx 


Consider  the  stress  in  any  member  of  the  truss  shown  at  (A), 
PI.  V,  as  C\DV  Taking  a  section  through  C^D^  C^F^  and  H^F^ 
the  origin  of  moments  for  computing  the  required  stress  (the 
moment-center)  may  be  taken  at  F2.  Let  vertical  lines  through 
Alt  Bv  Cv  Dl  intersect  the  funicular  polygon  in  A,  B,  C,  D, 
and  the  load  polygon  in  A',  B' ,  C',  D'.  Through  F2  draw  a 
vertical  line  intersecting  AB  in  G  and  CD  in  F\  then  (Art.  145) 
the  moment  of  the  required  stress  with  respect  to  the  origin  Fz 

M=HxFG, 

where  H  is  the  pole  distance  used  in  drawing  the  funicular 
polygon.  If  the  load  series  moves  (or  if  the  truss  is  considered 
as  moving  while  the  loads  remain  stationary),  the  stress  in  ClDl 
is  always  proportional  to  FG,  and  when  the  stress  is  a  maxi- 
mum or  a  minimum,  the  intercept  FG  is  also  a  maximum  or  a 
minimum. 

Let  the  vertical  through  Fl  intersect  A'B*  in  G'  and  C'D'  in 
F'.  It  will  now  be  shown  that,  as  the  truss  moves  horizontally, 
the  change  of  FG  (and  therefore  of  M)  per  unit  distance  moved 
is  always  proportional  to  F'G'.  The  reasoning  is  similar  to 
that  employed  in  Art.  131. 

Let  a,  b,  c,  d,  f,  g,  denote  the  ordinates  of  A,  B,  C,  D,  F,  G, 
respectively,  measured  from  the  horizontal  axis  LX;  and  a',  b', 
c',  d',f,  /,  the  ordinates  of  A',  B\  C,  D',  F',  G',  measured 
from  the  axis  L'X'.  Also,  let  z  denote  the  abscissa  of  the  point 
Alf  measured  from  the  axis  L'  V,  and  let  /j,  /2>  ^>  n\>  nz>  n  nave 
the  meanings  assigned  in  Art.  146. 

It  may  be  shown  as  in  Art.  131  that 


130 


GRAPHIC   STATICS. 


and  by  exactly  similar  reasoning, 


Therefore 


Similar  reasoning  applied  to  the  lines  A'  B' ,  CD1  leads  to  the 
equation 

The  rate  of  change  of  Mas  the  truss  moves  horizontally  is  found 
by  differentiating  (2)  with  respect  to  z.    Observing  (as  in  Art.  131) 

that  — ,  — ,  ...  are  the  values  of  —  at  A,  B,  .  . .,  equation  (i) 
dz  dz  dx 

shows  that 

~dz=0"1       ~dz=    ''"' 
the  differentiation  of  equation  (2)  therefore  gives 


dz      I          I          n 
Comparing  with  (3), 

*jL=g>-f  =  FG (4) 

It  follows  from  equation  (4)  that  when  the  position  of  the 
loads  is  such  that  M  is  a  maximum  or  a  minimum  (that  is,  when 
the  stress  in  C^D^  is  a  maximum  or  a  minimum),  the  ordinate 
F'G'  is  equal  to  zero. 

The  above  reasoning  is  general,*  and  the  conclusion  applies 
to  any  member  of  the  truss.  In  the  case  of  chord  members  the 
point  F±  will  in  general  fall  within  the  span ;  only  in  exceptional 

*  That  the  above  reasoning  is  rigorous  for  the  case  of  concentrated  loads  as  actually 
applied  to  the  truss  in  practice  follows  from  the  considerations  mentioned  in  the  note  in 
Art.  132.  With  slight  change  in  the  form  of  statement,  the  conclusion  holds  even  in  the 
ideal  case  of  concentrated  loads  in  the  strict  mathematical  sense. 


TRUSS  SUSTAINING  ANY  SERIES  OF  MOVING  LOADR      131 

forms  of  truss  will  it  fall  without  the  panel  C^D^  though 
coincide  with  £\  or  Dv  The  figure  for  the  case  of  a  web  mem- 
ber is  shown  at  (B\  PI.  V,  the  letters  A,  B,  C,  D,  F,  G,  with  or 
without  subscripts  or  accents,  denoting  corresponding  points  in 
the  constructions  shown  at  (A}  and  at  (.#).  The  point  F1  gener- 
ally falls  without  the  span  in  case  of  a  web  member;  it  is  possi- 
ble that  it  should  fall  between  Al  and  Cl  or  between  D1  and  Bv 
In  any  case  equation  (4)  holds,  and  the  condition  for  maximum 
or  minimum  stress  takes  the  above  general  form. 

The  criterion  above  reached  (F'G'=o)  is  easy  of  application. 
Let  a  strip  of  cardboard  be  prepared,  showing  the  truss-skele- 
ton and  the  moment-centers  for  all  members  of  the  truss.  By 
applying  this  to  the  diagram  showing  the  load  polygon,  the 
points  A'  and  B'  may  be  located  approximately  by  inspection 
(an  accurate  determination  of  them  not  usually  being  necessary 
for  the  purpose  in  hand);  by  stretching  a  thread  through  A1 
and  B',  the  point  G'  may  be  located  approximately ;  then,  hold- 
ing one  point  of  the  thread  at  G',  it  may  be  stretched  so  as  to 
pass  through  one  of  the  points  C'  and  D',  and  if  it  passes 
through  the  other  also,  the  condition  for  maximum  or  minimum 
stress  is  satisfied.  As  will  be  shown  presently,  when  the  load- 
series  consists  of  concentrated  loads,  a  load  will  generally  be  at 
Cl  or  D1  when  the  stress  is  a  maximum  ;  the  load  polygon  will 
therefore  be  vertical  (or  very  steep)  at  C'  or  at  D' ,  so  that  one 
of  these  points  may  be  shifted  vertically  without  changing 
appreciably  the  position  of  the  other  or  of  either  A'  or  B',  In 
some  cases  (but  not  usually)  it  may  be  necessary  to  locate  A', 
-B',  and  G'  accurately. 

150.  Positions  giving  Maximum  and  Minimum  Values  of  the 
Stress  Distinguished.  —  If  g—f  and  gl  — f  have  like  signs,  the 
magnitude  of  g—f  increases  with  a  positive  displacement,  and 
decreases  with  a  negative  displacement;  if  they  have  unlike 
signs,  g—f  decreases  with  a  positive  displacement,  and  increases 
with  a  negative  displacement.  It  follows  that,  if  the  truss  is  in 


132 


GRAPHIC   STATICS. 


a  position  of  maximum  stress  for  the  member  considered,  a 
small  positive  displacement  makes  the  signs  of  g—f  and  g'  — f 
unlike,  while  a  small  negative  displacement  makes  their  signs 
like ;  while  the  reverse  is  true  for  small  displacements  from  a 
position  of  minimum  stress.  (The  words  "  maximum "  and 
"  minimum  "  here  refer  to  magnitude,  irrespective  of  the  nature 
of  the  stress.) 

A  complete  discussion  of  this  principle  would  require  the 
consideration  of  various  forms  of  truss.  It  will  be  sufficient 
to  illustrate  the  application  to  the  form  shown  on  Plates  V 
and  VI. 

151.  Application  of  General  Principles.  —  On  PL  VI  is  repre- 
sented a  truss  of  60  ft.  span,  divided  into  six  equal  panels. 
The  greatest  depth  of  the  truss  is  12  ft. ;  the  lower  chord 
(carrying  the  moving  loads  at  the  joints)  is  straight,  while  the 
upper  joints  lie  upon  the  arc  of  a  circle.  The  two  web  mem- 
bers intersecting  at  any  upper  joint  are  of  equal  length  and 
slope. 

The  moving  load  consists  of  the  two  locomotives  and  train 
represented  on  PI.  Ill,  for  which  a  load  polygon  and  a  funicular 
polygon  are  drawn  on  PI.  VI.  The  diagram  represents  half  the 
actual  weight  of  the  train,  it  being  assumed  that  the  load  is 
divided  equally  between  two  trusses. 

The  construction  for  determining  the  greatest  stress  in  the 
chord  member  C1D1  is  shown  at  (A),  PI.  VI,  and  the  construc- 
tion for  the  case  of  the  web  member  HDl  is  shown  at  (£>).  Like 
lettering  is  used  in  the  two  cases. 

Member  of  lower  chord.  —  From  the  approximate  general  rule 
deduced  in  Art.  148,  it  follows  that,  in  order  that  the  stress  in 
ClDl  may  have  its  greatest  value,  the  truss  must  be  loaded  as 
completely  as  possible.  Moreover,  since  the  effect  of  a  given 
load  is  greater  the  nearer  it  is  to  the  panel  C-^Dlt  the  positions 
first  to  be  tested  are  those  bringing  the  heaviest  loads  upon  or 
near  the  panel. 


TRUSS  SUSTAINING  ANY  SERIES  OF  MOVING  LOADS. 


133 


The  condition  for  maximum  or  minimum  stress  is  found  to 
be  satisfied  in  the  position  in  which  the  truss  is  shown  in  the 
figure,  the  heavy  load  5  '6'  being  at  Dl  ;  but  not  in  any  other  posi- 
tion in  which  the  heaviest  loads  are  upon  or  near  the  panel  ClDl 
and  the  truss  fully  loaded.  The  criterion  of  Art.  150  shows 
conclusively  that  this  position  gives  a  maximum  and  not  a  mini- 
mum value  of  the  stress  ;  for  g—  f  is  positive,  and  a  positive 
displacement  of  the  truss  makes  g'  —f  negative,  while  a  nega- 
tive displacement  makes  it  positive.* 

The  value  of  the  maximum  stress  in  ClDl  may  now  be  deter- 
mined. The  pole  distance  is  100000  Ibs.  ;  FG=g.g  ft.;  F^F^ 
(the  moment-arm  of  the  stress)  is  12  ft.  The  required  stress  is 
therefore  a  tension  of 


Web  member.  —  Referring  to  (B\  PL  VI,  consider  the  mem- 
ber HDV  From  Art.  148  it  is  known  that  when  this  member 
sustains  its  greatest  tension  the  portion  D^B^  must  be  as  com- 
pletely loaded  as  possible  (with  A^C^  probably  free  from  loads), 
while  to  produce  the  greatest  compression  AlCl  must  be  as 
fully  loaded  as  possible  (with  D^B^  probably  free  from  loads). 
Whether,  in  either  case,  loads  should  be  between  Cl  and  D^  is 
not  known  without  applying  an  exact  test. 

Considering  the  case  of  tension,  let  the  load  series  be  brought 
from  the  right  until  the  foremost  load  reaches  the  panel,  and 
let  the  criterion  for  maximum  stress  be  applied  to  successive 
positions  until  one  is  found  in  which  it  is  satisfied.  It  is  found 
that  the  criterion  is  satisfied  when  the  second  load  (2-3)  is  at 
-Z^.  That  this  position  gives  a  maximum  and  not  a  minimum 
value  of  the  stress  is  found  by  applying  the  test  of  Art.  1  50. 
For  g—f  is  negative  ;  a  positive  displacement  of  the  truss 

*  It  will  be  noticed  that,  with  a  series  of  concentrated  loads,  F'G'  will,  as  a  rule,  pass 
through  zero  only  when  a  load  passes  one  of  the  four  points  Alt  Blt  Clt  D±.  Further, 
when  F'  is  between  C  and  D'  ',  a  load  at  A^  or  £1  (when  F'G'=o)  will  correspond  to  a 
minimum  stress,  while  a  load  at  Q  or  D±  will  correspond  to  a  maximum. 


134  GRAPHIC   STATICS. 

makes  g'-f  positive,  while  a  negative  displacement  makes  it 
negative.* 

The  value  of  the  maximum  tension  in  the  member  HDl  may 
now  be  found  by  the  method  of  moments.  The  intercept 
FG=8.8  ft;  the  pole  distance  is  100000  Ibs. ;  the  arm  F^J 
=  39.6  ft.  The  required  tension  is,  therefore,f 

8.8  x  i ooooo 


39-6 


=  22200  Ibs. 


152.   Algebraic  Statement  of  Condition  for  Maximum  or  Mini- 
mum Stress.  —  The  general  condition  above  deduced  admits  of 
simple  algebraic  expression.     For  this  purpose  let 
/>!  =  total  load  on  A^, 
Py=  total  load  on  D^, 
Q  =  total  load  on  C^D^ 
W=  total  load  on  A^^P^  +  P^+Q. 

Let  /j,  /2,  »j,  #2,  /,  n  have  the  meanings  assigned  in  Art.  146, 
their  algebraic  signs  being  carefully  observed. 

Referring  to  Fig.  (A\  PI.  V,  the  point  G'  divides  A'B'  into 
segments  proportional  to  /x  and  /2,  and  F'  divides  C'D'  into  seg- 
ments proportional  to  «j  and  «2.  Also,  with  the  notation  of 
Art.  149,  it  is  seen  that 

</-*'  =  />!  ;  b'-d!  =  Pz- 

d'-c'=Q;    b'-a'^W^P^  +  P^+Q. 

Since  A'  G'B'  is  a  straight  line, 

>-a>     V-<     V-a'      W 


*  Here  again  it  is  seen  that  the  condition  for  maximum  or  minimum  stress  will,  in 
general,  be  satisfied  only  when  a  load  is  at  one  of  the  four  points  A±,  B^,  C-^,  D±,  In  this 
case  a  maximum  stress  of  the  kind  now  under  consideration  will  occur  only  when  a  load 
is  at  DI. 

f  These  results,  being  obtained  from  small-scaled  drawings,  are  not  given  as  accurate 
values  of  the  stresses,  the  present  object  being  merely  to  illustrate  the  method  of  pro- 
cedure. By  careful  work,  and  with  drawings  made  to  a  suitable  scale,  a  sufficient  degree 
of  accuracy  may  be  obtained  by  the  above  method. 


TRUSS  SUSTAINING  ANY  SERIES  OF  MOVING  LOADS.      135 
hence  >-a'=™l    ............     l 


Similarly,  since  C'F'D'  is  a  straight  line, 


hence  /-,'  =      1;  ............  (3) 


Subtracting  equation  (3)  from  equation  (i),  and  (4)  from  (2), 
and  writing  P1  and  /^  instead  of  c'  —  a'  and  b'  —  d',  there  result 
the  equations  * 


f-S---P*  ..........  (6) 

If  the  points  T7'  and  G"'  coincide,  as  is  the  case  when  the  con- 
dition for  maximum  or  minimum  stress  is  satisfied,  g'  —  f'=o, 
and  equations  (5)  and  (6)  may  be  written  f 


Although  the  foregoing  discussion  has  referred  directly  to  the 

*  Equations  (5)  and  (6)  are  not  independent  ;  either  may  be  derived  from  the  other  by 
means  of  the  relations  PI  +  P%  +  (?  =  W,  /1  +  /2  =  /,  «1  +  «2  =  'z' 

t  It  is  interesting  to  compare  this  result  with  the  condition  for  maximum  or  minimum 
bending  moment  for  a  beam,  given  in  Art.  132.  If  the  portion  of  the  load  polygon  from 
C  to  D'  were  replaced  by  the  straight  line  C'  D'  (i.e.  if  the  load  Q  were  uniformly  dis- 

tributed over  the  panel),  the  total  load  from  A1  to  Fl  would  bePl  +  ~  Q,  and  the  total  load 

n 
from  F1  to  Bl  would  be  P2  +  —  Q.    The  condition  for  greatest  stress  in  C^Di  is  therefore 

identical  with  the  condition  for  maximum  bending  moment  at  Flt  on  the  assumption  that 
the  load  Q  is  uniformly  distributed.  This  interpretation  of  equation  (7)  is  of  use  only 
when  /*i  falls  between  Q  and  £>lt  although  it  holds  in  a  mathematical  sense  in  all  cases. 


1  36  GRAPHIC   STATICS. 

«ase  in  which  F1  falls  between  C±  and  Dlt  so  that  /v  /2,  «lf  «2  are 
all  positive,  the  result  holds  also  when  any  of  these  quantities 
are  negative,  signs  being  duly  observed  in  substituting  their 
values  in  equation  (7). 

If  the  member  considered  belongs  to  that  chord  to  which  the 
moving  loads  are  applied  (as  at  (A),  PL  V),  /j  and  /„  will  be 
positive,  and  one  or  both  the  quantities  »j  and  «2  will  be  posi- 
tive. If  one  of  the  members  C^F^  F^D^  were  vertical,  either 
»!  or  »a  would  be  zero.  In  no  truss  of  ordinary  design,  without 
subordinate  bracing,  would  n1  or  nz  be  negative  in  case  of  a 
chord  member. 

In  case  of  a  member  of  the  chord  carrying  no  moving  loads, 
the  point  Fl  falls  at  one  end  of  the  panel  C-J)-^  and  either  n-^  or 
n^  becomes  zero.  This  may  be  seen  by  referring  to  the  member 
Hjrv  PL  V,  (A). 

In  case  of  a  web  member  (as  C±H,  PL  V,  (/?)),  Fl  falls  without 
the  panel  ClDl  (making  n^  or  «2  negative)  ;  in  the  form  of  truss 
here  figured  F-^  also  falls  without  *  the  span  AlBl  (making  /x  or  /2 
negative).  In  this  case  equation  (7)  may  be  put  into  more  con- 
venient form.  Consider,  for  example,  the  maximum  tension  in 
the  member  D^H,  in  the  truss  shown  at  (B\  PL  VI.  From  the 
general  principles  stated  in  Art.  148  it  follows  that  loads  on 
DlBl  cause  tension  in  this  member,  while  loads  on  A1Cl  cause 
compression.  If  possible,  therefore,  equation  (7)  should  be 
satisfied  by  a  position  making  P1  zero  or  small.  The  equation 
may  be  written 


and  this  is  to  be  satisfied  by  a  position  making  Pt  =  o  or  as 
small  as  possible. 

Example.  —  Test  whether  the  general  equation  (7)  or  (8)  is 
satisfied  in  the  positions  shown  on  PL  VI. 

*  A  design  is  possible  which  will  cause  flt  for  certain  web  members,  to  fall  within  the 
span  but  without  the  panel  C\D\.    This  does  not  occur  in  any  ordinary  form  of  truss. 


TRUSS   WITH   SUBORDINATE   BRACING.  11    137**"  °F 

153.    Truss  with  Parallel  Chords. — The   application 
results  of  the  foregoing  discussion  to  the  case  of  trusses  with 
parallel  chords  presents  no  difficulty.*     The  only  matter  calling 
for  special  remark  relates  to  the  case  of  web  members. 

Referring  to  (B),  PL  VI,  it  will  be  seen  that  if  the  chords 
were  parallel  the  point  F1  (and  therefore  also  F  and  F1)  would 
fall  at  infinity.  In  order  that  C'D'  and  A'B'  might  intersect  in 
F',  these  two  lines  would  therefore  need  to  be  parallel.  The 
condition  for  maximum  stress  in  a  web  member  therefore 
reduces,  in  case  of  parallel  chords,  to  the  condition  that  A'B' 
and  CD'  shall  be  parallel. 

Further,  the  intercept  FG  and  the  moment-arm  F^J  become 
infinite  if  the  chords  are  parallel,  so  that  the  moment-equation 
takes  an  indeterminate  form.  The  stress  in  the  member  HDl 
can,  however,  be  easily  determined  by  resolving  vertically  all 
forces  acting  upon  the  portion  of  the  truss  to  the  left  of  a 
section  cutting  the  member  in  question  and  two  chord  members. 
See  Art.  142. 

The  algebraic  formula  expressing  the  condition  for  maximum 
stress  in  a  web  member  (equation  (8),  Art.  152)  reduces  to 
simpler  form  in  case  of  parallel  chords.  If  Fl  passes  to  infinity, 
/!  and  »1  approach  a  ratio  of  equality  while  —  approaches  zero ; 
so  that  the  equation  becomes,  in  the  limit, 


The  load  Pl  does  not  enter  this  equation,  but  is  to  be  taken  as 
small  as  possible  in  applying  it. 

§    4.    Truss  with  Subordinate  Bracing. 

154.  Description  of  Truss.  —  In  Fig.  49,  PI.  VIII,  is  repre- 
sented a  form  of  truss  designed  to  offer  points  of  support  for 
the  floor  system  intermediate  between  the  main  panel  joints. 

*  For  an  exhaustive  treatment  of  this  subject  consult  a  paper  by  Professor  Henry  T. 
Eddy,  Trans.  Am.  Soc.  C.  E.,  Vol.  XXII  (1890). 


I38  GRAPHIC    STATICS. 

The  effect  of  the  subordinate  vertical  members  a  and  diagonal 
members  b  requires  some  explanation.  With  the  arrangement 
shown  in  Fig.  49,  the  main  diagonals  are  tension  members  and 
the  main  verticals  compression  members,  while  the  subordinate 
verticals  sustain  tension  and  the  subordinate  diagonals  compres- 
sion. In  the  form  shown  in  Fig.  50  (PI.  VIII)  the  subordinate 
diagonals  sustain  tension.  The  members  represented  by  the 
broken  lines  in  both  Fig.  49  and  Fig.  50  do  not  act  when  the 
truss  sustains  dead  loads  only.  Such  members  will  in  general 
be  needed  only  in  panels  near  the  middle  of  the  span.  In 
Fig.  49  the  two  upper  diagonals  (and  in  Fig.  50  the  two  lower 
diagonals)  in  each  main  panel  are  designed  for  tension  only ; 
the  one  represented  by  the  broken  line  comes  into  action  only 
when  the  loading  is  such  that  the  other  would  be  thrown  into 
compression. 

155.  Determinateness  of  Stresses.  —  Disregarding  counter- 
braces,  it  is  easily  seen  that  the  stresses  in  all  members  are 
determinate. 

In  the  case  of  members  such  as  <?,  /,  or  g  (Fig.  49),  the  deter- 
minateness  follows  from  the  fact  that  any  such  member  can  be 
made  one  of  three  through  which  a  section  may  be  passed  com- 
pletely dividing  the  truss. 

The  stress  in  g1  is  determinate,  being  always  equal  to  that  in  g. 
For  a  like  reason,  the  stress  in  a  is  determinate,  being  always 
equal  to  the  load  carried  at  Cv 

Considering  the  point  of  intersection  of  #,  b,  d,  e,  since  the 
stresses  in  a  and  e  are  determinate,  it  follows  that  the  stresses 
in  b  and  d  are  also  determinate  ;  the  force  polygon  for  any  four 
forces  in  equilibrium  being  completely  determinate  when  two  of 
the  forces  are  known  completely,  and  the  directions  of  the  other 
two  are  given.  This  shows  that  the  stresses  in  b  and  d  would 
be  determinate  even  if  d  and  e  were  not  collinear. 

Like  reasoning  applied  to  an  upper  joint  shows  that  the 
stress  in  any  main  vertical  h  is  determinate. 


TRUSS   WITH    SUBORDINATE   BRACING. 


139 


156.  Effect  of  Subordinate  Braces  on  Stresses  in  Main  Mem- 
bers. —  Upper  chord.  —  It  may  be  shown  that  the  stresses  in  the 
upper  chord  members  have,  for  any  position  of  the  loads,  the 
same  values  they  would  have  if  the  subordinate  members  were 
not  present.  Thus,  applying  the  method  of  moments  in  the 
usual  manner,  the  center  of  moments  for  the  member  f*  is 
the  point  Dv  The  stress  is  determined  by  dividing  the  truss 
by  a  section  through/,  e,  and^,  and  applying  the  conditions  of 
equilibrium  to  either  portion  of  the  truss.  For  a  load  between 
A1  and  Fv  consider  the  right  portion  of  the  truss.  The  reaction 
at  Bl  is  the  same,  whatever  the  division  into  panels ;  hence  the 
stress  is  the  same  as  if  the  members  a  and  b  were  not  present. 
For  a  load  between  B^  and  Dv  the  same  conclusion  is  reached 
by  considering  the  left  portion  of  the  truss.  For  a  load  between 
Fl  and  D±  the  moment  equation  for  the  right  portion  of  the 
truss  will  contain  the  reaction  at  Bly  and  the  portion  of  the  load 
carried  at  Dl ;  and  this  portion  is  changed  by  the  subdivision 
of  the  panel  F-^D^  The  moment  of  the  load  at  Dl  is,  however, 
zero ;  so  that  the  subdivision  of  the  panel  has  no  effect  upon 
the  moment  equation  nor  upon  the  stress  in  the  member/. 

Web  members.  —  At  any  upper  joint  of  the  truss  four  forces  are 
in  equilibrium,  —  two  due  to  the  chord  stresses  and  two  to  the 
stresses  in  the  web  members.  Any  two  of  these  being  given, 
the  other  two  are  determined.  Since  the  chord  stresses  have 
the  same  value  whether  the  subordinate  members  are  present 
or  not,  the  same  is  true  of  the  web  stresses.  In  determining 
the  stresses  in  d  and  h,  the  subordinate  members  a  and  b  may 
therefore  be  disregarded. 

The  stress  in  e  is  changed  by  the  presence  of  the  subordinate 
members  if  any  load  is  between  Fl  and  D^.  The  value  of  the 
stress,  may,  however,  be  determined  by  the  method  of  sections, 
the  center  of  moments  being  at  the  intersection  of  /  and  g. 
The  construction  shown  at  (B\  PI.  VI,  may  be  applied  to  the 

*  The  present  discussion  refers  always  to  Fig.  49,  PI.  VIII. 


140 


GRAPHIC   STATICS. 


member  e,  the  short  panel  ClDl  (Fig.  49)  corresponding  to  the 
panel  ClDl  in  PL  VI,  and  the  point  F1  being  at  the  intersection 
of/  and  g. 

Lower  chord.  — The  analysis  just  applied  to  the  member  e  is 
applicable  also  to  a  lower  chord  member,  such  as  g.  Taking  a 
section  through/",  ^,  and^,  the  origin  of  moments  is  to  be  taken 
at  the  intersection  of  e  and  f.  The  construction  shown  on 
PI.  VI  may  therefore  be  applied,  the  panel  marked  ClDl  in 
Fig.  49  corresponding  to  C1Dl  in  PL  VI. 

Since  the  load  at  C±  and  the  member  a  are  collinear,  the 
stress  in  g1  is  always  equal  to  that  in  g. 

157.  Condition  for  Maximum  Stress  in  Main  Truss  Members. — 
From  the  results  of  Art.  156,  it  follows  that  the  method  already 
explained  for  determining  what  position  of  the  loads  causes  a 
maximum  stress  may  be  applied  to  all  the  main  members  of  the 
truss  shown  in  Fig.  49.  The  condition  in  all  cases  is  that  the 
points  F1  and  G',  determined  as  on  PL  V  or  PL  VI,  shall  coin- 
cide. In  case  of  certain  members,  however,  the  position  of  the 
line  CD'  is  changed  by  the  presence  of  the  subordinate  mem- 
bers. To  summarize  : 

In  the  case  of  the  members  /,  d,  and  h  (Fig.  49),  the  subordi- 
nate members  are  to  be  disregarded  in  applying  the  condition 
for  maximum  stress ;  the  whole  panel  FlDl  takes  the  place  of 
C^D^  in  the  construction  on  PL  VI.  This  principle  applies  to 
all  upper  chord  members,  all  main  verticals,  and  the  upper  por- 
tions of  all  main  diagonals. 

In  the  case  of  the  members  e  and  g,  the  short  panel  marked 
ClDl  in  Fig.  49  takes  the  place  of  C1D1  in  the  construction  on 
PL  VI ;  this  construction  being  otherwise  unchanged.  This 
applies  to  all  lower  chord  members,  and  to  the  lower  portions 
of  the  main  diagonals. 

There  is  no  difficulty  in  applying  similar  reasoning  to  the 
truss  shown  in  Fig.  50. 


TRUSS  WITH  SUBORDINATE  BRACING. 


141 


158.  Subordinate  Members.  —  When  the  position  of  the  load 
series  is  known,  the  stresses  in  the  subordinate  members  (a  and 
b,  Fig.  49)  are  easily  determined. 

The  stress  in  a  is  always  equal  to  the  load  carried  at  £\. 

The  members  d  and  e  being  collinear,  the  resolved  parts 
of  the  stresses  in  a  and  b  perpendicular  to  d  and  e  must  be 
equal ;  the  stress  in  b  can  therefore  be  found  directly  from  that 
in  a. 

It  remains  to  determine  the  position  of  the  load  series  which 
makes  the  load  at  C^  (and  therefore  the  stresses  in  a  and  b)  a 
maximum. 

Assuming  that  a  load  on  any  panel  is  carried  to  the  truss 
wholly  at  the  ends  of  that  panel,  the  condition  for  maximum 
load  at  any  joint  may  be  deduced  without  difficulty. 

In  Fig.  51  are  represented  the  load  polygon  and  funicular 
polygon  for  a  series  of  loads,  AlFl  and  FlBl  being  consecutive 


panels  of  any  lengths.  Vertical  lines  through  Alt  B^  and  P1 
intersect  the  funicular  polygon  in  A,  B,  F,  and  the  load  polygon 
in  A't  B' ,  F' .  The  portion  of  the  loads  on  AlFl  which  is  car- 


142 


GRAPHIC   STATICS. 


ried  at  F1  is  found  by  drawing  the  closing  string  AF,  and 
parallel  to  it  the  ray  OK' ;  the  portion  carried  at  Fl  is  K*2. 
Similarly,  drawing  OK"  parallel  to  the  closing  string  FB,  2  Kn 
represents  the  portion  of  the  loads  on  F1B1  which  is  carried  at 
Fr  The  total  load  carried  at  FL  is  therefore  K'K". 

Prolonging  BF  to  Z,  AFL  and  K'OK"  are  similar  triangles; 
hence  AL  and  K'K"  vary  in  the  same  ratio,  as  the  load  series 
moves.  But  if  G  is  the  point  in  which  a  vertical  through  F 
intersects  AB,  it  is  seen  that  FG  varies  as  AL.  Therefore, 
when  K'K"  is  a  maximum,  FG  is  a  maximum. 

Reasoning  as  in  Arts.  131  and  132,  it  is  found  that  when  FG 
is  a  maximum  or  a  minimum  the  points  A',  F!,  and  B'  are 
collinear. 

The  condition  for  greatest  load  at  FL  is  therefore  exactly 
analogous  to  the  condition  for  greatest  bending  moment  at  FL 
on  the  supposition  that  A^B^  is  a  simple  beam  supported  at 
A1  and  Bv 

From  this  analogy  the  conditions  for  greatest  stresses  in  the 
members  a  and  b  (Fig.  49)  may  be  stated  as  follows :  * 

(a)  The  panels  FlCl  and  C-J)-^  must  be  loaded  as  heavily  as 
possible,  the  heaviest  loads  being  near  Cv 

(6)  The  loads  on  the  panels  must  be  proportional  to  their 
lengths. 

In  satisfying  requirement  (b\  a  load  must  generally  be  at  Cv 
and  must  be  regarded  as  divided  between  the  panels  in  some 
arbitrary  manner. 

159.  Effect  of  Counterbraces.  —  The  foregoing  discussion  and 
conclusions  need  modification  when  the  loading  is  such  as  to 
bring  counterbraces  into  action.  It  is  needful  to  determine 
what  counters  are  needed,  what  maximum  stresses  they  sustain, 
and  whether  the  maximum  stresses  in  any  other  members  are 
changed  by  their  action.  Let  it  be  assumed  that  when  the 

*See  paper  by  Professor  H.  T.  Eddy,  Trans.  Am.  Soc.  C.  E.,  Vol.  XXII. 


TRUSS   WITH   SUBORDINATE   BRACING. 


143 


train  moves  from  right  to  left  the  members  called  into  action 
are  those  represented  in  Fig.  52.  The  conclusions  drawn  in 
Art.  155  as  to  the  determinateness  of  the  stresses  are  equally 
applicable  to  the  present  case ;  but  the  fact  that  the  mem- 
bers d  and  e  are  not  collinear  makes  it  necessary  to  modify 


Fig.  52 

the  methods  of  determining  greatest  stresses  in  certain 
members. 

In  case  of  the  members  /,  g,  e,  and  a  (Fig.  52)  the  usual 
methods  obviously  apply. 

If  b  is  designed  for  tension  only,  it  is  necessary  to  determine 
whether  it  acts  when  the  stress  in  d  is  greatest.  If  it  does  not, 
the  discussion  of  d  is  obvious.  If  b  does  act,  the  stress  in  d  is 
not  the  same  as  in  the  case  in  which  d  and  e  are  collinear.  It 
may  be  shown,  however,  that  the  stress  in  d  always  bears  a 
constant  ratio  to  that  which  would  exist  if  d  and  e  were  collinear. 
This  becomes  evident  if  the  force  polygon  be  drawn  for  the 
forces  due  to  the  stresses  in  f,  f,  d,  and  h,  and  the  effect  of 
changing  the  direction  of  d  be  observed ;  it  being  remembered 
that  such  change  does  not  affect  the  stresses  mf,f. 

In  a  truss  of  long  span,  the  stress  in  d  will  usually  be  small, 
since  the  dead  load  will  be  so  great  as  to  nearly  or  quite  coun- 
terbalance the  greatest  live  load  effect.  Counters  will  be  needed, 
if  at  all,  only  in  panels  near  the  middle  of  the  span. 

The  determination  of  the  maximum  stress  in  b  with  the 
counter  d  in  action  is  less  simple  than  it  would  be  if  d  and 
e  were  collinear;  but  the  member  b  will  have  its  true  maxi- 
mum when  the  member  d  does  not  act,  the  train  approaching 
from  the  left. 


144 


GRAPHIC   STATICS. 


1 60.  Influence  Lines.  —  From  the  foregoing  discussion  it  is 
seen  that  the  influence  lines  for  the  main  truss  members  may  be 
drawn  by  the  general  method  given  in  Art  147.     In  one  case, 
however,  there  is  a  peculiarity  which  is  worthy  of  special  notice. 

Considering  the  lower  chord  member  C1D1  (Fig.  49,  PL  VIII), 
the  point  Fl  falls  between  Al  and  Clt  making  ;/j  negative.  The 

ordinate  F2f2  is  made  equal  to  -y^,  positive ;  A2f2  and  f^B^  are 

drawn,  intersecting  verticals  through  C^  and  Z>2  in  CB  and  D3, 
respectively;  the  influence  line  is  A2C3DS£2.  This  peculiar 
form  does  not  occur  in  any  common  form  of  truss  without 
subordinate  bracing. 

Figure  50  shows  the  influence  line  for  the  upper  chord  mem- 
ber g ;  the  subordinate  braces  being  so  placed  that  the  diago- 
nals are  in  tension.  Here  »2  is  negative,  Fl  falling  between 
Dl and  Bv 

For  the  subordinate  members  the  influence  lines  may  be 
drawn  without  difficulty  from  the  principles  of  Art.  158. 

161.  Application. — On  PL  VII  is  shown  a  truss  of  300  ft. 
span,  divided  into  ten  panels  of  30  ft.  each.     The  lengths  of  the 
main  verticals  are  30  ft.,  45  ft.,  and  60  ft.     The  subordinate 
diagonals  are  so  placed  as  to  sustain  tension.     Counterbraces 
are  shown  in  the  central  panel  only;    a  full  computation  of 
stresses  due  to  live  and  dead  loads  will  determine  whether  other 
counters  are  needed.     The  moving  load  is  the  same  as  shown 
on  PL  III. 

At  (A\  PL  VII,  is  shown  the  position  of  the  truss  causing 
greatest  stress  in  the  upper  chord  member  C^F^,  and  at  (B)  the 
position  causing  greatest  stress  in  the  web  member  C-^D^.  The 
letters  A,  B,  C,  D,  F  have  the  same  meanings  as  on  Plates  V 
and  VI,  and  the  construction  is  self-explanatory.  If  the  lower 
part  of  the  web  member  CJK,  in  Fig.  (B\  were  under  consid- 
eration, the  point  F-^  would  be  as  shown  in  the  figure,  but  C1D1 
would  be  the  double  panel  C^K  (the  same  as  if  the  subordinate 


UNIFORMLY   DISTRIBUTED   MOVING   LOAD.  145 

members  were  omitted).  These  statements  are  in  accordance 
with  the  foregoing  discussion. 

In  each  of  the  cases  shown  on  PL  VII  it  will  be  seen  that  the 
lines  A'B'  and  C'D1  intersect  at  a  point  F'  lying  in  the  vertical 
through  Fv 

The  stresses  corresponding  to  the  positions  thus  determined 
may  be  computed  by  the  method  of  moments,  as  on  PL  VI. 
The  construction  is  not  shown ;  but  satisfactory  results  may  be 
secured  by  careful  work  and  by  the  use  of  suitable  linear  and 
force  scales. 

The  construction  for  determining  the  position  for  greatest 
stress  in  a  subordinate  member  is  not  shown ;  it  is  identical  with 
the  construction  shown  on  PL  IV  for  determining  the  position 
for  greatest  bending  moment  at  a  given  section  of  a  beam. 


§  5.    Uniformly  Distributed  Moving  Load. 

162.  Application  of  General  Method.  —  The  case  in  which  the 
moving  load  is  assumed  to  have  a  uniform  horizontal  distribu- 
tion involves  no  principles  additional  to  those  already  explained. 
The  foregoing  discussion  applies  to  any  load  distribution  what- 
ever. It  is,  however,  of  interest  to  notice  the  special  form 
assumed  by  the  general  principles  when  the  load  is  uniformly 
distributed. 

The  load  polygon  becomes  a  straight  line  whose  slope  depends 
upon  the  intensity  of  the  loading  (the  load  per  unit  length). 
The  funicular  polygon  becomes  a  continuous  curve,  being  in 
fact  a  parabola  with  axis  vertical,  the  position  of  the  vertex 
depending  upon  the  position  of  the  pole  of  the  force  diagram. 
For  the  purpose  of  drawing  the  funicular  polygon  the  load  may 
be  replaced  by  a  series  of  concentrated  loads ;  the  sides  of  the 
resulting  polygon  will  be  tangent  to  the  desired  curve,  and  the 
polygon  will  coincide  more  and  more  nearly  with  the  desired 
parabola,  the  smaller  and  nearer  together  the  concentrated  loads 
are  taken. 


I46  GRAPHIC   STATICS. 

It  is  obvious  that  for  maximum  chord  stresses  the  only  condi- 
tion is  that  the  moving  load  shall  cover  the  whole  span. 

For  maximum  stress  in  any  web  member,  the  head  of  the 
load  series  must  be  at  some  point  in  the  corresponding  panel ; 
the  exact  position  may  be  determined  by  applying  the  general 
method.  The  position  of  the  head  of  the  load  series  is  shown 
at  a  glance  by  the  influence  diagram.  Thus,  from  diagram  (C), 
Fig.  48  (PI.  VIII),  it  is  obvious  that  the  greatest  tension  in  the 
member  C^Dl  will  occur  when  the  load  covers  ElBl  completely, 
E^Ai  being  free  from  load. 

163.  Assumption  of  Full  Panel  Loads.  —  The  determination 
of  the  maximum  web  stresses  is  somewhat  simplified  by  an 
approximate  assumption  as  to  the  way  in  which  the  uniformly 
distributed  load  comes  upon  the  truss.  Thus,  in  Fig.  53,  let  it 
be  required  to  determine  the  greatest  tension  in  the  member 
BB' .  The  moving  load  must  be  brought  on  from  the  right  until 
its  head  is  at  some  point  between  A  and  B.  In  this  position, 
the  loads  actually  supported  at  the  various  joints  are  as  follows: 

A'  B'          c' D' 


X.  A  BC"CZD  E  F  Y 

Pig.  53 

At  C  a  full  panel  load  (half  the  load  between  B  and  C,  and  half 
that  between  C  and  D} ;  at  each  of  the  points  D,  E,  and  F  also 
a  full  panel  load  ;  at  B  less  than  a  full  panel  load  (half  the  load 
on  BC,  and  a  portion  of  that  on  AE).  Let  it  be  assumed  that, 
when  the  tension  in  BB'  is  greatest,  all  joints  from  B  to  /'inclu- 
sive sustain  full  panel  loads,  while  all  joints  to  the  left  of  B 
(only  one  in  the  case  shown)  are  free  from  loads. 

If  a  similar  assumption  is  made  in  case  of  every  chord  mem- 
ber, the  resulting  stresses  will  not  differ  greatly  from  those 
obtained  by  rigorously  adhering  to  the  assumption  of  a  uniformly 
distributed  load,  the  error  being  on  the  side  of  safety.  It  is  to 


UNIFORMLY   DISTRIBUTED   MOVING   LOAD.  147 

be  remarked,  also,  that  the  case  of  a  strictly  uniform  dis- 
tribution is  not  realized  in  practice,  so  that  the  results  of 
the  above  assumption  are  probably  as  reliable  as  would  be 
obtained  by  following  out  strictly  the  assumption  of  uniform 
distribution. 

The  method  of  determining  maximum  stresses,  on  the  above 
assumption  as  to  the  loading,  will  be  illustrated  by  reference  to 
the  truss  shown  in  Fig.  54,  PI.  VIII.  In  this  truss  the  panels 
are  of  equal  length ;  but  the  method  applies  also  to  the  case  of 
unequal  panels,  the  only  difference  being  that  the  "  full  panel 
loads  "  for  different  joints  are  not  equal  unless  the  panels  are 
equal. 

The  principle  of  counterbracing  will  be  employed  here,  the 
diagonals  being  constructed  to  sustain  tension  only. 

164.  Dead  Load  Stresses. — The  dead  load  stresses  may 
be  determined  by  means  of  a  stress  diagram,  as  in  the  roof- 
truss  problems  already  treated.  Two  points  should  be  ob- 
served in  drawing  this  diagram,  (i)  If  dead  loads  are  taken 
to  act  at  upper  as  well  as  at  lower  joints,  the  force  poly- 
gon for  the  loads  and  reactions  must  show  these  forces  in 
the  same  order  as  that  in  which  their  points  of  application 
occur  in  the  perimeter  of  the  truss.  (2)  The  diagonal  mem- 
bers assumed  to  act  are  taken  as  all  sloping  in  the  same 
direction. 

The  reason  for  the  first  point  is  the  same  as  already  explained 
in  Art.  90;  viz.  that  unless  the  forces  be  taken  in  the  order 
mentioned,  the  stress  diagram  cannot  be  the  true  reciprocal  of 
the  truss  diagram  and  certain  lines  will  have  to  be  duplicated. 
The  reason  for  assuming  the  diagonals  as  all  sloping  in  the 
same  way  is  the  same  as  in  the  case  of  the  roof  truss  with 
counterbracing  (Arts,  in  and  112). 

The  dead-load  stress  diagram  is  not  shown,  since  its  con- 
struction involves  no  principle  not  already  fully  explained  and 
illustrated. 


I48  GRAPHIC    STATICS. 

165.  Stresses  Due  to  Moving  Load. — (a)  Chord  members. — 
By  applying  the  method  of  sections  it  is  easily  seen  that  a  load 
at  any  point  produces  tension  in  every  lower  chord  member  and 
compression  in  every  upper  chord  member.  It  follows  that  the 
greatest  stresses  in  the  chord  members  will  occur  when  the  truss 
is  fully  loaded.  A  convenient  method  of  determining  the  live 
load  stresses  is,  therefore,  to  draw  a  stress  diagram,  as  in  case  of 
fixed  loads.  This  diagram  is  not  shown. 

(6)  Web  members.  —  In  case  of  a  web  member,  loads  in  dif- 
ferent positions  tend  to  cause  opposite  kinds  of  stress.  Thus, 
considering  the  member  f'n'  (Fig.  54,  PI.  VIII),  a  tension  is 
caused  in  it  by  a  load  at  either  of  the  points  ab,  be,  cd,  de,  or  ef, 
while  a  compression  is  caused  by  a  load  at  fg,  gh,  or  hi.  Simi- 
larly, loads  at  ab,  be,  cd,  de,  and  ef  all  tend  to  throw  compression 
on  the  vertical  member  f'm',  while  loads  on  fg,  gh,  hi  have  the 
opposite  tendency. 

Therefore,  to  produce  the  greatest  tension  upon  fn'  (and 
•compression  on  f'm'),  the  live  load  must  act  only  at  ef  and  all 
joints  to  the  right;  while  to  cause  the  greatest  compression  on 
f'n'  (and  tension  upon  fm')  the  live'load  must  act  only  at  fg, 
gh,  and  hi.  A  similar  statement  will  hold  regarding  any  other 
web  member.  Since  counterbraces  are  to  be  used  in  all  panels 
in  which  diagonal  members  would  otherwise  be  thrown  into 
compression,  we  shall  need  only  to  consider  the  greatest  tension 
in  each  diagonal  and  the  greatest  compression  in  each  vertical. 
We  shall  first  outline  the  method  to  be  employed,  and  then 
explain  the  construction. 

To  determine  the  greatest  tension  in  a  diagonal  member,  as 
g'm' :  Assume  the  live  load  to  come  upon  the  bridge  from  the 
right  until  there  are  full  loads  at  the  joints  ab,  be,  cd,  de,  ef,  and 
fg.  Take  a  section  cutting  g'm'  and  the  two  chord  members 
gg'  and  m'm,  and  consider  the  forces  acting  upon  the  portion  of 
the  truss  to  the  left  of  the  section.  These  forces  are  four  in 
number :  the  reaction  at  the  support  and  the  forces  acting  in  the 
three  members  cut.  Hence  we  first  determine  the  reaction, 


UNIFORMLY   DISTRIBUTED   MOVING   LOAD. 


and  then  determine  the  three  other  forces  for  equilibrium 
the  method  of  Art.  42. 

The  construction  is  shown  in  Fig.  54  (PI.  VIII).  ABCDEF 
is  the  force  polygon  for  the  eight  live  loads  that  may  come 
upon  the  truss.  Choosing  a  pole  O,  the  funicular  polygon 
for  the  eight  loads  is  next  drawn.  Now,  turning  the  attention 
to  the  member  g'm',  the  loads  gh  and  /«'  are  assumed  not  to  act. 
The  reactions  at  the  supports  for  this  case  of  loading  are  found 
in  the  usual  way.  Prolong  oa  and  og  to  intersect  the  lines 
of  action  of  the  two  reactions,  and  join  the  two  points  thus 
determined.  This  gives  the  closing  line  of  the  funicular  poly- 
gon (or  om\  The  ray  OMis  now  drawn  parallel  to  the  string 
om,  and  the  two  reactions  are  GM  and  MA. 

Now  take  a  section  through  mm'  ,  m'g1,  and  g'g)  and  apply 
the  construction  of  Art.  42  to  the  determination  of  the  forces 
acting  in  the  three  members  cut.  The  resultant  of  GM  and 
the  force  in.gg'  must  act  through  the  point  X  (the  intersection 
of  gg  produced  and  «;').  The  resultant  of  the  forces  in  mmr 
and  m'g*  must  act  through  their  intersection  Y.  Hence  these 
two  resultants  (being  in  equilibrium  with  each  other)  must  both 
act  in  the  line  XY.  From  G  draw  a  line  parallel  to  gg1  ,  and 
from  Ma.  line  parallel  to  XY;  mark  their  point  of  intersection 
G'.  Then  MG'  is  the  resultant  of  the  forces  in  mm'  and  m'g'. 
From  M  draw  a  line  parallel  to  mm',  and  from  G'  a  line  parallel 
to  the  member  g'm'  ,  and  mark  their  point  of  intersection  M'. 
Then  M'  G'  represents  the  force  in  the  member  m'g1  .  This  is 
also  the  value  of  the  greatest  stress  in  m'g'  . 

To  determine  the  stress  in  the  vertical  member  I'g1,  the 
same  loading  must  be  assumed,  and  a  similar  construction  is 
employed.  Take  a  section  cutting  //',  I'g3,  and  g'g,  and  deter- 
mine forces  acting  in  these  three  lines  which  shall  be  in  equi- 
librium with  the  reaction  GM.  This  reaction  is  in  equilibrium 
with  MG'  and  G'G,  the  former  having  the  line  of  action  XY. 
Then  MG'  is  resolved  into  two  forces  having  the  directions  of 
the  members  g'l'  and  /'/.  The  stress  in  g'l'  is  found  to  be 


ISO 


GRAPHIC   STATICS. 


a  compression,  represented  in  the  stress  diagram  by  the  line 
G'L'. 

The  maximum  live  load  stress  in  every  web  member  may  be 
found  in  the  same  way.  If  the  above  reasoning  is  understood, 
there  will  be  no  difficulty  in  applying  the  same  method  to  the 
remaining  members. 

1 66.  Maximum  Stresses.  —  By  combining  the  stresses  due  to 
live  and  dead  loads,  the  maximum  stresses  are  easily  deter- 
mined. 

Web  members.  —  When  the  web  members  are  considered,  the 
effect  of  counterbracing  needs  careful  attention. 

The  construction  above  explained  gives  the  greatest  live  load 
tension  in  each  diagonal.  It  may  be  that  for  certain  members, 
this  tension  is  wholly  counterbalanced  by  the  dead  loads.  In 
any  panel  in  which  this  is  the  case,  the  member  shown  will 
never  act  and  may  be  omitted.  The  counterbrace  must  then 
be  considered. 

Evidently,  the  algebraic  sum  of  the  stresses  due  to  live  and 
dead  loads  will  be  the  true  maximum  tension  in  each  of  the 
diagonals  shown.  For  the  greatest  tension  in  the  other  system 
of  diagonals,  the  load  must  be  brought  on  the  bridge  from  the 
left;  or,  what  amounts  to  the  same  thing,  the  tension  already 
found  for  any  one  of  the  diagonals  shown  is  also  the  greatest 
tension  in  the  diagonal  sloping  the  opposite  way  in  the  panel 
equally  distant  from  the  middle  of  the  truss.  (In  fact,  the 
stresses  in  all  members  due  to  a  movement  of  the  loads  from 
left  to  right  may  be  obtained  from  the  results  already  reached, 
by  consideration  of  symmetry.) 

As  to  the  vertical  members,  two  values  of  the  stress  must  be 
compared  in  every  case  — -  namely,  the  greatest  compressions 
corresponding  to  the  two  directions  of  the  moving  load.  But 
both  can  be  obtained  from  the  above  results  by  considering  the 
symmetry  of  the  truss.  For  example,  the  stress  found  for  I'g* 
is  a  possible  value  for  the  stress  in  /£',  and  must  be  compared 


UNIFORMLY   DISTRIBUTED   MOVING   LOAD.  151 

with  the  value  obtained  for  the  latter  member  when  the  load 
moves  from  right  to  left.  It  is  possible,  also,  that  certain  of 
the  verticals  may  be  in  tension  when  the  dead  loads  act 
alone. 

Maximum  chord  stresses. — These  are  found  by  combining 
the  stresses  due  to  fixed  and  moving  loads,  determined  as 
already  described. 


PART   III. 
CENTROIDS  AND  MOMENTS  OF  INERTIA 


CHAPTER   VIII.      CENTROIDS. 
§  i .    Centroid  of  Parallel  Forces. 

167.  Composition  of  Parallel  Forces.  —  The  composition   of 
complanar  parallel  forces   can   always   be   effected   by  means 
of  the  funicular  polygon,  by  the  method  of  Art.  27.     It  is  rjow 
necessary  to  consider  parallel  systems  more  at  length,  as  a  pre- 
liminary to  the  discussion  of  graphic  methods  for  determining 
centers  of  gravity  and  moments  of  inertia. 

1 68.  Resultant  of  Two  Parallel  Forces.  —  Let  ab  and  be  (Fig. 
55)  be  the  lines  of  action  of  two  parallel  forces,  their  magni- 
tudes being  AB  and  BC  (not  shown).     Let 
ac  be  the  line  of  action  of  their  resultant, 
and  AC  (not  shown)   its    magnitude.       By 
the  principle  of  moments  (Art.  50)  the  sum 
of  the  moments  of  AB  and  BC  about  any 
point  in  their  plane  is  equal  to  the  moment 

of  AC  about  the  same  point.  If  the  origin  of  moments  is  on 
ac,  the  moment  of  AC  is  zero;  and  therefore  the  moments  of 
AB  and  BC  are  numerically  equal  (but  of  opposite  signs). 

Let  any  line  be  drawn  perpendicular  to  the  given  forces, 
intersecting  their  lines  of  action  in  /",  Q' ,  and  R'  respectively. 
Then  AB  x  P'R'  =  BCx  Q'R1. 
152 


"Fig.  05 


CENTROID  OF  PARALLEL  FORCES. 


153 


Let  any  other  line  be  drawn  intersecting  the  three  lines  of 
action  in  P,  Q,  and  R  respectively.  Then 

PR  =  QR 
P'R'     Q'R'1 

and  therefore  AB  x  PR  =  BCx  QR. 

That  is,  PQ  is  divided  by  the  line  ac  into  segments  inversely 

proportional  to  AB  and  BC. 

If  AB  and  BC  act  in  opposite  directions,  the  line  ac  will  be 
outside  the  space  included  between  ab  and  be ;  but  the  above 
result  is  true  for  either  case. 

169.  Centroid  of  Two  Parallel  Forces.  — If  the  lines  of  action 
ab  and  be  (Fig.  55)  be  turned  through  any  angle  about  the  points 
P  and   Q  respectively,   the  forces   remaining  parallel -and   of 
unchanged  magnitudes,  the  line  of  action  of  their  resultant  will 
always  pass  through  the  point  R.     For,  by  the  preceding  article, 
the  line  of  action  of  the  resultant  will  always  intersect  PQ  in  a 
point  which  divides  PQ  into  segments  inversely  proportional  to 
AB  and  BC.     Hence,  if  AB  and  BC  remain  unchanged,  and 
also  the  points  P  and  Q,  the  point  R  must  also  remain  fixed. 

If  P  and  Q  are  taken  as  the  points  of  application  of  AB  and 
BC,  R  may  be  taken  as  the  point  of  application  of  AC,  in 
whatever  direction  the  parallel  forces  are  supposed  to  act.  The 
point  R  is  called  the  centroid  *  of  the  parallel  forces  AB  and 
BC  for  the  fixed  points  of  application  P  and  Q. 

1 70.  Centroid  of  Any  Number  of  Parallel  Forces.  —  Let  AB, 

BC,  and  CD  be  three  parallel  forces,  and  let  P,  Q,  and  5 
(Fig.  56)  be  their  fixed  points  of 
application.  Let  R  be  the  cen- 
troid of  AB  and  BC,  and  AC  their 
resultant.  Take  R  as  the  fixed 
point  of  application  of  AC,  and 
determine  T,  the  centroid  of.  AC  Fig.se 

*  The  name  center  of  parallel  forces  has  been  quite  commonly  used  instead  of  centroid 
as  above  defined.  The  latter  term  has,  however,  been  adopted  by  some  of  the  later  writers, 
and  seems  on  the  whole  a  better  designation. 


54 


GRAPHIC   STATICS. 


and  CD.  Let  AD  be  the  resultant  of  AC  and  CD ;  then  AD  is 
also  the  resultant  of  AB,  BC,  and  CD. 

Now  if  AB,  BC,  and  CD  have  their  direction  changed,  but 
still  remain  parallel  and  unchanged  in  magnitude,  it  is  evident 
that  the  point  T,  determined  as  above,  will  remain  fixed  and  will 
always  be  on  ad,  the  line  of  action  of  AD.  The  point  T  is  called 
the  centroid  of  the  three  forces  AB,  BC,  and  CD. 

By  an  extension  of  the  same  method,  a  centroid  may  be 
determined  for  any  system  of  parallel  forces  having  fixed  points 
of  application.  Hence  the  following  definition  may  be  stated  : 

The  centroid  of  a  system  of  parallel  forces  having  fixed  points 
of  application  is  a  point  through  which  the  line  of  action  of 
their  resultant  passes,  in  whatever  direction  the  forces  be  taken 
to  act. 

In  determining  the  centroid  by  the  method  just  described, 
the  forces  may  be  taken  in  any  order  without  changing  the 
result.  For  the  centroid  must  lie  on  the  line  of  action  of  the 
resultant ;  and  since  this  is  a  determinate  line  for  each  direction 
in  which  the  forces  may  be  taken  to  act,  there  can  be  but  one 
centroid. 

171.  Non-complanar  Parallel  Forces.  —  The  reasoning  of  the 
preceding  articles  is  equally  true,  whether  the  forces  are  corn- 
planar   or   not.      In   what   follows   we   shall   deal   either   with 
complanar  forces,  or  with  forces  whose  points  of  application  are 
complanar.     No  more  general  case  will  be  discussed. 

172.  Graphic  Determination  of  Centroid  of  Parallel  Forces. — 
If  the  line  of  action  of  the  resultant  of  any  system  of  parallel 
forces  be  found  for  each  of  two  assumed  directions  of  the  forces, 
the  point  of  intersection  of  these  two  lines  is  the  centroid  of 
the  system.     Moreover,  if  the  points  of  application  are  com- 
planar, the  two  assumed  directions  may  both  be  such  that  the 
forces  will  be  complanar. 

Thus,  let  the  plane  of  the  paper  be  the  plane  containing  the 
given  points  of  application,  and  let  ab,  be,  cd,  de,  ef  (Fig.  57)  be 


CENTROID  OF  PARALLEL  FORCES. 


155 


the  points  of  application  of  five  parallel  forces,  AB,  BC,  CD, 
DE,  EF.  Draw  through  these  points  parallel  lines  in  some 
chosen  direction,  and  taking  them  as  the  lines  of  action  of  the 
given  forces,  construct  the  force  and  funicular  polygons  corre- 


spending.  The  line  of  action  of  the  resultant  is  drawn  through 
the  intersection  of  the  strings  oa  and  of,  and  this  line  contains 
the  centroid  of  the  given  forces.  Next  draw  through  the  given 
points  of  application  another  set  of  parallel  lines,  preferably  per- 
pendicular to  the  set  first  drawn,  and  draw  a  funicular  polygon 
for  the  given  forces  with  these  lines  of  action.  (It  is  unneces- 
sary to  draw  a  new  force  diagram,  since  the  strings  in  the 
second  funicular  polygon  may  be  drawn  respectively  perpendic- 
ular to  those  of  the  first.)  This  construction  determines  a 
second  line  as  the  line  of  action  of  the  resultant  corresponding 
to  the  second  direction  of  the  forces.  The  required  centroid  of 
the  system  is  the  point  of  intersection  of  the  two  lines  of 
action  of  the  resultant  thus  determined,  and  is  the  point  af  in 
the  figure. 

Example.  —  Find  graphically  the  centroid  of  the  following 
system  of  parallel  forces,  and  test  the  result  by  algebraic  com- 
putation :  A  force  of  20  Ibs.  applied  at  a  point  whose  rectangu- 
lar coordinates  are  (4,  6) ;  12  Ibs.  at  the  point  (12,  3) ;  20  Ibs.  at 


156  GRAPHIC   STATICS. 

the  point  (10,  10)  ;  — 10  Ibs.  at  the  point  (7,  —9);  —8  Ibs.  at 
the  point  (—5,  —  10). 

173.  Centroid  of  a  Couple.  —  If  AB  and  BC  are  the  magni- 
tudes of  any  two  parallel  forces  applied  at  points  P  and   Q, 
then  by  Art.  169  their  centroid  R  is  on  the  line  PQ  and  is 
determined  by  the  equation 

P_R  =  BC_ 
QR     AB' 

If  AB  and  BC  are  equal  and  opposite  forces,  PR  and  QR  must 
be  numerically  equal,  and  R  must  be  outside  the  space  between 
the  lines  of  action  of  AB  and  BC.  These  conditions  can  be 
satisfied  only  by  making  PR  and  QR  infinite.  Hence  we  may 
say  that  the  centroid  of  two  equal  and  opposite  forces  lies  on 
the  line  joining  their  points  of  application  and  is  infinitely  dis- 
tant from  these  points. 

174.  Centroid  of  a  System   whose  Resultant  is  a  Couple. — 

If  a  system  with  fixed  points  of  application  is  equivalent  to  a 
couple,  its  centroid  will  be  infinitely  distant  from  the  given 
points  of  application.  A  line  containing  this  centroid  can  be 
determined  as  follows : 

Take  the  given  forces  in  two  groups ;  the  resultants  of  the 
two  groups  will  be  equal  and  opposite.  Find  the  centroid  of 
each  group,  and  suppose  each  partial  resultant  applied  at  the 
corresponding  centroid.  Then  the  centroid  of  the  whole  sys- 
tem is  the  same  as  that  of  the  couple  thus  formed,  and  will  lie 
on  the  line  joining  the  two  partial  centroids. 

If  the  separation  into  groups  be  made  in  different  ways,  dif- 
ferent couples  and  different  partial  centroids  will  be  found. 
The  different  couples  are,  of  course,  equivalent ;  and  it  may  be 
proved  that  the  lines  joining  the  different  pairs  of  partial  cen- 
troids are  all  parallel,  and  intersect  in  the  (infinitely  distant) 
centroid  of  the  whole  system. 

For,  suppose  the  given  system  to  be  equivalent  to  a  couple 


CENTER   OF   GRAVITY. 

Q  with  points  of  application  A  and  B,  and  also  t 

with  points  of  application  A'  and  B'.     These  two 

be  equivalent  to  each  other,  whatever  be  the  directio 

forces.     Let  AB  be  taken  as  this  direction.     The  two  equal 

and  opposite  forces  of  the  couple  Q  neutralize  each  other,  since 

their  lines  of  action  are  coincident ;  hence  the  two  forces  of 

the  couple  Q'  must  also  neutralize  each  other.     Therefore  A'B' 

must  be  parallel  to  AB. 

175.  Moment  of  a  Force  about  an  Axis.  —  The  moment  of  a 
force  with  respect  to  a  given  axis,  as  defined  in  Art.  47, 
depends  not  only  upon  the  point  of  application  of  the  force, 
but  upon  its  direction.  In  dealing  with  systems  of  forces 
whose  direction  may  change  but  whose  points  of  application 
are  complanar,  we  shall  need  to  compute  moments  only  for 
axes  lying  in  the  plane  of  the  points  of  application ;  and  the 
forces  may  usually  be  regarded  as  acting  in  lines  perpendicular 
to  this  plane.  Hence  we  shall  compute  moments  by  the 
following  rule  : 

The  moment  of  a  force  with  reference  to  any  axis  is  the 
product  of  the  magnitude  of  the  force  into  the  distance  of  its 
point  of  application  from  the  axis. 


§  2.    Center  of  Gravity  —  Definitions  and  General  Principles. 

176.  Center  of  Gravity  of  Any  Body.  —  Every  particle  of 
a  terrestrial  body  is  attracted  by  the  earth  with  a  force  propor- 
tional directly  to  the  mass  of  the  particle.  The  resultant  of 
such  forces  upon  all  the  particles  of  a  body  is  its  weight ;  and 
the  point  of  application  of  this  resultant  is  called  the  center  of 
gravity  of  the  body.  The  lines  of  action  of  these  forces  may 
be  assumed  parallel  without  appreciable  error.  We  may  there- 
fore define  the  center  of  gravity  of  a  body  as  follows : 

If  forces  be  supposed  to  act  in  the  same  direction  upon  all 
particles  of  a  body,  each  force  being  proportional  to  the  mass 


158  GRAPHIC   STATICS. 

of  the  particle  upon  which  it  acts,  the  centroid  of  this  system 
of  parallel  forces  is  the  center  of  gravity  of  the  body. 

This  point  is  also  called  center  of  mass,  and  center  of  inertia, 
either  of  which  is  a  better  designation  than  center  of  gravity. 
The  latter  term  is,  however,  in  more  general  use. 

177.  Centers   of  Gravity  of  Areas   and  Lines. — The  term 
center  of    gravity,   as   above  defined,   has   no   meaning  when 
applied   to   lines  and  areas,   since   these  magnitudes  have  no 
mass,  and  hence  are  not  acted  upon  by  the  force  of  gravity. 
It  is,  however,  common  to  use  the  name  center  of  gravity  in 
the  case  of  lines  and  areas,  with  meanings  which  may  be  stated 
as  follows  : 

The  center  of  gravity  of  an  area  is  the  center  of  gravity  of 
its  mass,  on  the  supposition  that  each  superficial  element  has  a 
mass  proportional  to  its  area.  This  point  would  be  better 
described  as  the  center  of  area. 

The  center  of  gravity  of  a  line  is  the  center  of  gravity  of  its 
mass,  on  the  assumption  that  each  linear  element  has  a  mass 
proportional  to  its  length.  The  term  center  of  length  is  pref- 
erable, and  will  often  be  used  in  what  follows. 

Similar  statements  might  be  made  regarding  geometrical 
solids,  but  we  shall  have  to  deal  chiefly  with  lines  and  areas. 

178.  Moments  of  Areas  and   Lines. — Definition. — The  mo- 
ment of  a  plane  area  with  reference  to  an  axis  lying  in  its  plane 
is  the  product  of    the  area  by  the  distance  of  its  center  of 
gravity  from  the  axis. 

Proposition.  —  The  moment  of  any  area  about  a  given  axis  is 
equal  to  the  sum  of  the  moments  of  any  set  of  partial  areas 
into  which  it  may  be  divided.  For,  by  the  definition  of  center 
of  gravity,  a  force  numerically  equal  to  the  total  area  and 
applied  at  its  center  of  gravity  is  the  resultant  of  a  system  of 
forces  numerically  equal  to  the  partial  areas  and  applied  at 
r.rjir  respective  centers  of  gravity  ;  and  the  moment  of  any 


CENTER   OF   GRAVITY.  159 

force  is  equal  to  the  sum  of  the  moments  of  its  components 
(Art  50). 

A  similar  definition  and  proposition  may  be  stated  regarding 
lines. 

The  moment  of  an  area  or  line  is  zero  for  any  axis  containing 
its  center  of  gravity. 

179.  Symmetry.  —  Two   points   are   symmetrically  situated 
with  respect  to  a  third  point  if  the  line  joining  them  is  bisected 
by  that  point. 

Two  points  are  symmetrically  situated  with  respect  to  a  line 
or  plane  when  the  line  joining  them  is  perpendicular  to  the 
given  line  or  plane  and  bisected  by  it. 

A  body  is  symmetrical  with  respect  to  a  point,  a  line,  or  a 
plane,  if  for  every  point  in  the  body  there  is  another  such  that 
the  two  are  symmetrically  situated  with  respect  to  the  given 
point,  line,  or  plane.  The  point,  line,  or  plane  is  in  this  case 
called  a  point  of  symmetry,  an  axis  of  symmetry,  or  a  plane  of 
symmetry  of  the  body. 

180.  General  Principles.  —  (i)  The  center  of  gravity  of  two 
masses  taken  together  is  on   the  line  joining  the  centers  of 
gravity  of  the  separate  masses.     For  it  is  the  point  of  applica- 
tion of    the  resultant  of  two  parallel  forces  applied  at  those 
points. 

(2)  If  a  body  of  uniform  density  has  a  plane  of  symmetry, 
the  center  of  gravity  lies  in  this  plane.  If  there  is  an  axis  of 
symmetry,  the  center  of  gravity  lies  in  this  axis.  If  the  body 
is  symmetrical  with  respect  to  a  point,  that  point  is  the  center 
of  gravity.  For  the  elementary  portions  of  the  body  may  be 
taken  in  pairs  such  that  for  each  pair  the  center  of  gravity  is 
in  the  plane,  axis,  or  point  of  symmetry. 

181.  Centroid. — The  center  of  gravity  of  any  body  or  geo- 
metrical magnitude  is  by  definition  the  same  as  the  centroid  of 
a   certain    system    of   parallel   forces.     It   will    be 


160  GRAPHIC    STATICS. 

therefore,  to  use  the  word  centroid  in  most  cases  instead  of 
center  of  gravity. 

§  3.    Centroids  of  Lines  and  of  Areas. 

182.  General   Method   of   Finding   Centroid.  —  The  centroid 
of  any  area  may  be  found  by  the  following  method  :   Divide  the 
given  area  into  parts  such  that  the  area  and  centroid  of  each 
part  are  known.     Take  the  centroids  of  the  partial  areas  as  the 
points  of  application  of  forces  proportional  respectively  to  those 
areas.     The  centroid  of  this  system  of  forces  is  the  centroid  of 
the  total  area,  and  may  be  found  by  the  method  of  Art.  172. 

The  centroid  of  a  line  may  be  found  by  a  similar  method. 

The  method  just  described  is  exact  if  the  magnitudes  of  the 
partial  areas  and  their  centroids  are  accurately  known.  If  the 
given  area  is  such  that  it  cannot  be  divided  into  known  parts,  it 
will  still  be  possible  to  get  an  approximate  result  by  this 
method. 

In  applying  this  general  method,  it  is  frequently  necessary  to 
know  the  centroids  of  certain  geometrical  lines  and  figures,  and 
also  the  relative  magnitudes  of  the  areas  of  such  figures. 
Methods  of  determining  these  will  be  given  in  the  following 
articles. 

183.  Centroids  of  Lines. — The  centroid  of  a  straight  line  is 
at  its  middle  point. 

Broken  line.  —  The  centroid  of  a  broken  line  is  the  center  of 
a  system  of  parallel  forces,  of  magnitudes  proportional  to  the 
lengths  of  the  straight  portions  of  the  line,  and  applied  respec- 
tively at  their  middle  points.  It  may  be  found  graphically  by 
the  method  of  Art.  172,  or  by  any  other  method  applicable  to 
parallel  forces. 

Part  of  regular  polygon.  —  For  the  centroid  of  a  part  of  a 
regular  polygon,  a  special  construction  is  found  useful. 


CENTROIDS    OF    LINES    AND    OF   AREAS.  161 

Let  ABCDE  (Fig.  58)  be  part  of  such  a  polygon,  and  O  the 
center  of  the  inscribed  circle.  Let 
r  =  radius  of  inscribed  circle  ;  /  = 
length  of  a  side  of  the  polygon  ;  s 
=  total  length  of  the  broken  line 
AE.  Through  O  draw  OC,  the  axis 


M  A'Q         °  E'N        of  symmetry  of  AE;  and  MN,  per- 

Fig.  as  T     i  /-i/- 

pendicular  to  OC. 

First,  the  centroid  must  lie  on  OC. 

Second,  to  find  its  distance  from  O,  assume  a  system  of  equal 
and  parallel  forces  applied  at  the  middle  points  of  the  sides  AB, 
BC,  etc.  The  required  centroid  is  the  point  of  application  of 
the  resultant  of  these  forces.  Taking  MN  as  the  axis  of 
moments,  and  letting  x  =  required  distance  of  centroid  from 
MN,  and  x^,  xz,  xz,  x±  the  distances  of  the  middle  points  of  AB, 
BC,  etc.,  from  MN,  we  have  from  the  principle  of  moments, 


But  if  Ab  and  Bb  be  drawn  perpendicular  respectively  to  PQ 
and  J/TVwe  have  from  the  similar  triangles  ABb,  POQ, 

AB_=PO_      J_  =  L 
Ab~ PQ™  Ab~xi 


In  the  same  way, 

Ix2=r*  be, 


Hence,  s-x=r  (Ab  +  bc+cd+dE)  =  r- AE, 

where  AE  is  equal  to  the  projection  of  the  broken  line  ABCDE 
on  MN. 

The  centroid  G  may  now  be  found  graphically  as  follows  : 
Make  Oc'  =  r\  ON=%s\  OE1  =  \  AE;  draw  Nc' .  Then  G  is 
determined  by  drawing  E' G  parallel  to  Nc' 


l62  GRAPHIC    STATICS. 

Circular  arc. — The  above  construction  holds,  whatever  the 
.length  of  the  side  /.  If  this  length  be  decreased  indefinitely, 
while  the  number  of  sides  is  increased  indefinitely,  so  that  the 
length  s  remains  finite,  we  reach  as  the  limiting  case  a  circular 
arc.  The  same  construction  therefore  applies  to  the  determina- 
tion of  the  centroid  of  such  an  arc,  r  denoting  the  radius  of  the 
circle  and  s  the  length  of  the  arc. 

184.  Centroids  of  Geometrical  Areas.  —  Parallelogram.  —  The 
centroid  of  a  parallelogram  is  on  a  line  bisecting  two  opposite 
sides. 

Let  ABCD  (Fig.  59)  be  a  parallelogram,  and  EF  a  line 
bisecting  AD  and  BC.  Divide  AB  into  any  even  number  of 
equal  parts,  and  through  the  points 
of  division  draw  lines  parallel  to  BC. 
Also  divide  J5Cinto  any  even  number 
of  equal  parts  and  draw  through  the 
points  of  division  lines  parallel  to 
AB.  The  given  parallelogram  is  thus 
divided  into  equal  elements.  Now 
consider  a  pair  of  these  elements,  such  as  those  marked  P  and 
Q  in  the  figure,  equally  distant  from  AD,  and  also  equally 
distant  from  EF,  but  on  opposite  sides  of  it.  The  centroid  of 
the  two  elements  taken  together  is  at  the  middle  point  of  the 
line  joining  their  separate  centroids.  If  the  number  of  divisions 
of  AB  and  of  BC  be  increased  without  limit,  the  elements 
approach  zero  in  area,  and  the  centroids  of  P  and  Q  evidently 
approach  two  points  which  are  equally  distant  from  EF.  Hence 
in  the  limit,  the  centroid  of  such  a  pair  of  elements  lies  on  the 
line  EF.  But  the  whole  area  ABCD  is  made  up  of  such  pairs  ; 
hence  the  centroid  of  the  whole  area  is  on  the  line  EF.  For 
like  reasons  it  is  also  on  the  line  bisecting  AB  and  DC]  hence 
it  is  at  the  intersection  of  the  two  bisectors. 

The  point  thus  determined  evidently  coincides  with  the  point 
of  intersection  of  the  diagonals  AC  and  BD. 


CENTROIDS   OF   LINES   AND   OF   AREAS,  rg, 

u.  o*  £• 

Triangle.  —  The  centroid  of  a  triangle  lies  on  a  lil 
from  any  vertex  to  the  middle  of  the  opposite  side ; 
therefore,  the  point  of  intersection  of  the  three  such  lines. 

Let  ABC  (Fig.  60)  be  any  triangle,  and  D  the  middle  point 
of  BC.     Then  the  centroid  of  ABC  must  lie  on  AD.     For  AD 
bisects  all  lines,  such  as  be,  parallel  to  BC.     Now  inscribe  in 
the  triangle  any  number  of  parallelograms  such  as  bcc'b',  with 
sides  parallel  respectively  to  BC  and 
AD.     The  centroid  of  each  parallelo- 
gram lies  on  AD,  and,  therefore,  so 
also  does  the  centroid  of  the  whole 
area  composed  of  such  parallelograms. 
If  the  number  of  such  parallelograms 
be  increased  without  limit,  the  alti- 
tude of  each  being  diminished  without  limit,   their  combined 
area  will  approach  that  of  the  triangle,  and  the  centroid  of  this 
area  will  approach  in  position  that  of  the  triangle.     But  since 
the  former  point  is  always  on  the  line  AD,  its  limiting  position 
must  be  on  that  line.     Therefore  the  line  AD  contains  the  cen- 
troid of  the  triangle. 

By  the  same  reasoning,  it  follows  that  the  centroid  of  ABC 
must  lie  on  BE,  drawn  from  B  to  the  middle  point  of  AC. 
Hence  it  must  be  the  point  of  intersection  of  AD  and  BE, 
which  point  must  also  lie  on  the  line  CF  drawn  from  C  to  the 
middle  point  of  AB. 

The  point  G  divides  each  bisector  into  segments  which  are 
to  each  other  as  i  to  2.  For,  from  the  similar  triangles  ABC, 
EDC,  since  EC  is  half  of  AC,  it  follows  that  DE  is  equal  to 
half  of  BA.  And  from  the  similar  triangles  AGB,  DGE,  since 
DE  is  half  of  AB,  it  follows  that  GE  is  half  of  GB,  and  GD 
half  of  GA. 

Quadrilateral.  —  Let  ABCD  (Fig.  61)  be  a  quadrilateral  of 
which  it  is  required  to  find  the  center  of  gravity.  Draw  BD, 
and  let  E  be  its  middle  point.  Make  EGi=^  EA,  and  EG2  =  \ 
EC.  Then  the  centroids  of  the  triangles  ABD  and  BCD  are 


1 64 


GRAPHIC   STATICS. 


G!  and  G2  respectively.     Hence  the  centroid  of  ABCD  is  on 

the  line  G-iGi  at  a  point  dividing  it  into  segments  inversely 
proportional  to  the  areas  of  ABD 
and  BCD.  Since  these  two  tri- 
angles have  a  common  base,  'their 
areas  are  proportional  to  their  alti- 
tudes measured  from  this  base. 
But  these  altitudes  are  propor- 

tional to  AF  and  FC,  or  to  G^H  and  G^H  ;  hence,  if  G  is  the 

required  centroid, 


Fig.  61 


Therefore  G  is  found  by  making  G^G=  GZH. 

Circular  sector.  —  To  find  the  centroid  of  a  circular  sector 
OAB  (Fig.  62)  we  may  reason  as  follows  :  Draw  two  radii  OM, 
ON,  very  near  together.  Then  OMN  differs  little  from  a  tri- 
angle, and  its  centroid  will  fall  very  near  the  arc  A'B',  drawn 
with  radius  equal  to  two-thirds  of  OA.  If  the  whole  sector  be 
subdivided  into  elements  such  as  OMN,  their  centers  of  gravity 
will  all  fall  very  near  to  the  arc  A'B'.  If  the 
number  of  such  elements  is  indefinitely  in- 
creased, the  line  joining  their  centroids 
approaches  as  a  limit  the  arc  A'B'.  And 
since  the  areas  of  the  elements  are  propor- 
tional to  the  lengths  of  the  corresponding 
portions  of  A'B',  the  centroid  of  the  total 
area  is  the  same  as  that  of  the  arc  A'B'. 
This  point  may  be  found  by  the  method  described  in  Art.  183. 


M  N 


185.  Graphic  Determination  of  Areas.  —  Let  there  be  given 
any  number  of  geometrical  figures,  and  let  it  be  required  to 
determine  the  relative  magnitudes  of  their  areas. 

If  a  number  of  rectangles  can  be  found,  of  areas  equal 
respectively  to  the  areas  of  the  given  figures  and  having  one 
common  side,  then  the  remaining  sides  will  be  proportional  to 
the  areas  of  the  given  figures. 


CENTROIDS   OF   LINES   AND   OF   AREAS.  165 

An  important  case  is  that  in  which  the  given  figures  are 
such  that  the  area  of  each  is  equal  to  the  product  of  two  known 
lines.  In  this  case  a  series  of  equivalent  rectangles  having  one 
common  side  'can  be  found  by  the  following  construction. 

Let  ABC  (Fig.  63)  be  a  triangle,  and  let  it  be  required  to 
determine  an  equivalent 
rectangle  having  a  side  of 
given  length  as  LM.  Let 
b  and  k  be  the  base  and 
altitude  of  the  triangle. 
MakeZ7^=i  b  and  LP  =  h, 
and  draw  the  semicircumference  PRN.  Draw  the  ordinate  LR 
perpendicular  to  PN  ";  then 

bh  =  area  ABC. 


Draw  MR,  and  perpendicular  to  it  draw  RQ.     Then 

ABC. 


Hence  LQ  is  the  required  length. 

If  the  given  figure  is  a  parallelogram,  LN  and  LP  may  be 
its  base  and  altitude.  If  it  be  a  circular  sector,  LN  and  LP 
may  be  the  length  of  the  arc  and  half  the  radius. 

1  86.  Centroids  of  Partial  Areas.  —  It  may  be  required  to 
find  the  centroid  of  the  part  remaining  after  deducting  known 
parts  from  a  given  area.  For  this  case  the  construction  of. 
Art.  182  needs  modification.  In  the  case  there  considered  we 
regarded  the  partial  areas  and  the  total  area  as  proportional  to 
parallel  forces  acting  at  their  respective  centers  of  gravity.  In 
this  case  we  may  also  represent  the  total  area,  the  portions 
deducted  from  it,  and  the  remaining  portion  as  forces  acting  at 
the  respective  centers  of  gravity  ;  but  the  forces  corresponding 
to  the  areas  deducted  must  be  taken  as  acting  in  the  opposite 
direction  to  that  assumed  for  the  forces  representing  the  total 
area  and  the  area  remaining. 

Thus,  to  find  the  centroid  of  the  area  remaining  after  deduct- 


166  GRAPHIC   STATICS. 

.kkg.,irom  the  circle  ABD  (Fig.  64)  a  smaller  circle  EFH  and  a 
sector  OAB,  we  may  proceed  as  follows  :  Find  the  centroid  of 
three  parallel  forces  proportional  to  the 
areas  ABD>  EFH,  and  OAB,  and  applied 
at  their  respective  centroids  O,  C,  and  G; 
but  the  last  two  must  be  taken  as  acting 
in  the  direction  opposite  to  that  of  the  first. 
With  this  understanding,  the  force  and 
funicular  polygons  may  be  employed  as  in 
Art.  172. 

187.  Moments  of  Areas.  — The  moment  of  an  area  about  any 
line  in  its  plane  may  be  determined  from  the  funicular  polygon 
employed  in  finding  its  center  of  gravity.  Let  the  parallel 
forces  applied  at  the  centroids  of  the  partial  areas  be  assumed 
to  act  parallel  to  the  axis  of  moments.  Then  the  distance  inter- 
cepted on  the  axis  by  the  extreme  lines  of  the  funicular  poly- 
gon, multiplied  by  the  pole  distance,  is  equal  to  the  moment  of 
the  total  area  about  the  axis.  For,  by  Art.  56,  this  construc- 
tion gives  the  moment  of  the  resultant  force  about  any  point 
in  the  given  axis ;  and  this  is  equal  to  the  moment  of  the  result- 
ant area  about  the  axis  by  definition. 

A  similar  rule  gives  the  moments  of  the  partial  areas. 


t).  O*  C. 


CHAPTER    IX.       MOMENTS    OF    INE^IA. 
§    I.    Moments  of  Inertia  of  Forces. 

1 88.  Definitions. — The  moment  of  inertia  of  a  body  with 
respect  to  an  axis  is  the  sum  of  the  products  obtained  by  mul- 
tiplying the  mass  of  every  elementary  portion  of  the  body  by 
the  square  of  its  distance  from  the  given  axis. 

The  moment  of  inertia  of  an  area  with  respect  to  an  axis  is 
the  sum  of  the  products  obtained  by  multiplying  each  element- 
ary area  by  the  square  of  its  distance  from  the  axis. 

The  moment  of  inertia  of  a  line  may  be  similarly  denned, 
using  elements  of  length  instead  of  elements  of  mass  or  area. 

The  moment  of  inertia  of  a  force  with  respect  to  any  axis  is 
the  product  of  the  magnitude  of  the  force  into  the  square  of  the 
distance  of  its  point  of  application  from  the  axis.  The  sum  of 
such  products  for  any  system  of  parallel  forces  is  the  moment 
of  inertia  of  the  system  with  respect  to  the  given  axis. 

[NOTE.  —  The  term  moment  of  inertia  had  reference  originally  to  material  bodies, 
the  quantity  thus  designated  having  especial  significance  in  dynamical  problems  relat- 
ing to  the  rotation  of  rigid  bodies.  The  quantity  above  defined  as  the  moment  of 
inertia  of  an  area  is  of  frequent  occurrence  in  the  discussion  of  beams,  columns,  and 
shafts  in  the  mechanics  of  materials.  In  the  graphic  discussion  of  moments  of  iner- 
tia of  areas,  it  is  convenient  to  treat  areas  as  forces,  just  as  in  the  determination  of 
centers  of  gravity;  it  is  therefore  convenient  to  use  the  term  moment  of  inertia  of  a 
force  in  the  sense  above  defined.  It  is  only  in  the  case  of  masses  that  the  term 
moment  of  inertia  is  really  appropriate,  but  it  is  by  analogy  convenient  to  apply  it 
to  the  other  cases.] 

The  product  of  inertia  of  a  mass  with  respect  to  two  planes  is 
the  sum  of  the  products  obtained  by  multiplying  each  element- 
ary mass  by  the  product  of  its  distances  from  the  two  planes. 

167 


168  GRAPHIC    STATICS, 

The  product  of  inertia  of  an  area,  a  line,  or  a  force  may  be 
defined  in  a  similar  manner. 

For  a  plane.t  area,  the  product  of  inertia  with  respect  to 
two  lines  in  its  plane  may  be  defined  as  the  sum  of  the 
products  obtained  by  multiplying  each  element  of  area  by  the 
product  of  its  distances  from  the  given  lines.  This  is  equiv- 
alent to  the  product  of  inertia  of  the  area  with  respect  to  two 
planes  perpendicular  to  the  area  and  containing  the  two  given 
lines. 

For  a  system  of  forces  with  points  of  application  in  the  same 
plane,  the  product  of  inertia  with  reference  to  two  axes  in  that 
plane  may  be  defined  as  the  sum  of  the  products  obtained  by 
multiplying  the  magnitude  of  each  force  by  the  product  of  the 
distances  of  its  point  of  application  from  the  given  axes.  It  is 
with  such  systems  and  with  plane  areas  that  the  following  pages 
chiefly  deal. 

The  radius  of  gyration  of  a  body  with  respect  to  an  axis  is  the 
distance  from  the  axis  of  a  point  at  which,  if  the  whole  mass  of 
the  body  were  concentrated,  its  moment  of  inertia  would  be 
unchanged.  The  square  of  the  radius  of  gyration  is  equal  to 
the  quotient  obtained  by  dividing  the  moment  of  inertia  of  the 
body  by  its  mass. 

The  radius  of  gyration  of  an  area  may  be  defined  in  a  similar 
manner. 

The  radius  of  gyration  of  a  system  of  parallel  forces  is 
the  distance  from  the  axis  of  the  point  at  which  a  force 
equal  in  magnitude  to  their  resultant  must  act  in  order  that 
its  moment  of  inertia  may  be  the  same  as  that  of  the  system. 
The  square  of  the  radius  of  gyration  may  be  found  by  dividing 
the  moment  of  inertia  of  the  system  by  the  resultant  of  the 
forces. 

189.  Algebraic  Expressions  for  Moment  and  Product  of  In- 
ertia.—  Moment  of  inertia.  —  Let  m\,  m2,  etc.,  represent  ele- 
mentary masses  of  a  body,  and  r\,  ;•<>,  etc.,  their  respective 


MOMENTS   OF   INERTIA   OF   FORCES.  169 

distances  from  an  axis  ;  then  the  moment  of  inertia  of  the  body 
with  respect  to  that  axis  is 

m  \r\-\-  m^-\  —  =  2  m  r2, 

the  symbol  2  being  a  sign  of  summation,  and  the  second  mem- 
ber of  the  equation  being  merely  an  abbreviated  expression  for 
the  first. 

Product  of  inertia.  —  Let/i,  /2,  etc.,  denote  the  perpendicular 
distances  of  elements  mi,  m2,  etc.,  from  one  plane,  and  q^  q2,  etc., 
their  distances  from  another  ;  then  the  product  of  inertia  of  the 
body  with  respect  to  the  two  planes  is 

mipiqi  +  m2p2q2  H  —  =  2mpq. 

Radius  of  gyration.  —  With  the  same  notation,  if  k  denotes 
the  radius  of  gyration  of  the  body,  we  have 

(mi  +  m2-\  —  )  t?  =  miJ'i2  +  m&*  H  —  . 
Or, 


Here  2m  is  equal  to  the  whole  mass  of  the  body. 

Product-radius.  —  Let  c  represent  a  quantity  defined  by  the 
equation. 

(mi  +  m2-\  —  )  c2  =  mipiqi  +  7^2/2^2  H  —  • 

Then  if  the  whole  mass  were  concentrated  at  the  same  distance 
c  from  two  axes,  its  product  of  inertia  with  respect  to  those 
axes  would  be  unchanged.  This  quantity  c  is  thus  seen  to  be 
analogous  to  the  radius  of  gyration.  It  may  be  called  the 
product-radius  of  the  body  with  respect  to  the  two  axes.  The 
value  of  c  is  always  given  by  the  equation 

«Jtmpq 
2m  ' 

Expressions  similar  to  those  just  given  apply  also  to  plane 
areas  and  to  systems  of  parallel  forces.  In  case  of  an  area, 
mi,  m.2,  etc.,  denote  elements  of  area  ;  rlf  r2,  etc.,  their  distances 
from  the  axis  of  inertia;  and  (pi,  q^,  (pz,  g2),  etc.,  their  dis- 


IjO  GRAPHIC    STATICS. 

tances  from  the  two  planes.  In  case  of  a  system  of  parallel 
forces  with  complanar  points  of  application,  m^  m2,  etc.,  must 
be  replaced  by  the  magnitudes  of  the  forces. 

190.  Determination  of  Moment  of  Inertia  of  a  System  of 
Parallel  Forces.  —  Let  the  points  of  application  of  the  forces 
be  in  the  same  plane,  which  also  contains  the  assumed  axis. 
We  shall  have  to  deal  only  with  systems  satisfying  these  condi- 
tions. By  the  definition,  the  moment  of  inertia  will  be  the 
same,  whatever  the  direction  of  the  forces.  If  we  take  the 
moment  of  any  force  (as  defined  in  Art.  175)  about  the  given 
axis,  and  suppose  a  force  equal  in  magnitude  to  this  moment 
to  act  at  the  point  of  application  of  the  original  force,  and  in  a 
direction  corresponding  to  the  sign  of  the  moment,  then  the 
moment  of  this  new  force  about  the  given  axis  is  equal  to 
the  moment  of  inertia  of  the  original  force.  If  this  be  done  for 
all  the  forces,  the  algebraic  sum  of  the  results  will  be  the 
required  moment  of  inertia  of  the  system. 

This  process  can  be  carried  out  graphically  by  methods 
already  described. 

Let  ab,  be,  cd,  de  (Fig.  65)  be  the  points  of  application  of 
four  parallel  forces,  and  let  the  axis  of  inertia  be  QR.  Suppose 
all  the  forces  to  act  in  lines  parallel  to  QR,  passing  through  the 
given  points  of  application.  Their  respective  moments  with 
reference  to  QR  may  now  be  found  by  the  method  of  Art.  55. 

Draw  the  force  polygon  ABCDE  and  choose  a  pole  O,  taking 
the  pole  distance  H  preferably  equal  to  AE  or  some  simple 
multiple  of  AE.  (In  Fig.  65  H  is  taken  equal  to  AE.}  Draw 
a  funicular  polygon  and  prolong  each  string  to  intersect  QR. 
'Then  the  moment  of  any  force  with  respect  to  QR  is  the 
product  of  H  by  the  distance  intercepted  on  QR  by  the  two 
strings  corresponding  to  the  force  in  question.  Thus  the 
moment  of  AB  is  given  by  the  distance  A'B'  (Fig.  65)  multi- 
plied by  H.  Also  the  successive  moments  of  BC,  CD,  DE  are 
represented  by  £'C',  C'D',  D'E',  each  multiplied  by  H.  It  is 


MOMENTS    OF    INERTIA    OF   FORCES. 


171 


seen  also  that  the  intercepts,  if  read  in  the  above  order,  give  a 
distinction  'between  positive  and  negative  moments ;  upward 
distances  on  QR  denoting  in  this  case  positive  moments,  and 
downward  distances  negative  moments. 

We  have  now  to  find  the  sum  of  the  moments  of  a*  second 
system  of  forces  acting  in  the  original  lines,  but  represented  in 


Fig.  6S 

magnitude  and  direction  by  the  intercepts  just  found.  We 
may  take  as  the  force  polygon  for  the  second  construction  the 
line  A'B'C'D'E'  (Fig.  65),  and  choosing  any  pole  distance  H!, 
draw  a  second  funicular  polygon  and  find  the  distances  inter- 
cepted by  the  successive  strings  on  the  line  QR.  But  in  this 
case,  since  only  the  resultant  is  desired,  we  need  only  find  the 
intercept  A"E"  between  the  first  and  last  strings.  The  product 
of  this  intercept  by  H'  gives  the  sum  of  the  moments  of  A'B', 
B'C',  C'D',  D!E'  with  respect  to  QR  ;  and,  if  the  product  be 
multiplied  by  H  (since  A'B',  B'C',  etc.,  should  each  be  multi- 
plied by  H  in  order  to  represent  the  magnitudes  of  the  forces 
of  the  second  system),  the  result  will  be  the  required  moment  of 
inertia  of  the  given  system  of  forces  AB,  BC,  CD,  DE. 


i;2  GRAPHIC   STATICS. 

It  should  be  noticed  that  in  Fig.  65  (A)  is  a  force  diagram, 
(B)  a  space  diagram  (Art.  11);  that  is,  every  line  in  (A)  repre- 
sents a  force,  while  every  line  in  (B}  represents  a  distance. 
Even  A'B',  B'C',  etc.,  though  used  as  forces,  are  actually 
merely  distances ;  and  the  moment  of  any  one  of  them  is  the 
product  of  a  length  by  a  length. 

191.  Radius  of   Gyration.  —  The  moment  of   inertia  of   the 
given  system  is  HxH'  xA"E".     If  H  has  been  taken  equal 
to  AE,  the  product  H'xA"E"  must  equal  the  square  of  the 
radius  of  gyration  of  the  system  with  respect  to   QR.      The 
length  of  the  radius  of  gyration  can  be  found  as  follows  (Fig. 
65) :  Draw  LN=H'+A"E",  and  make  LM=H'.     On  LNas 
a  diameter  draw  a  semicircle  LPN,  and  from  M  draw  a  line 
perpendicular  to  LN,  intersecting  the  semicircle  in  P.     Then 
MP  is  the  length  of  the  required  radius  of  gyration.     For  by 
elementary  geometry  we  have  PMz  =  LMxMJV. 

If  H  is  taken  equal  to  n  X  AE,  the  moment  of  inertia  is  equal 
to  H'  xA"J£"  xnx  AE,  and  the  square  of  the  radius  of  gyration 
is  equal  to  nH'xA"E".  Hence  in  Fig.  65  we  should  put 
either  LM=nH',  or  MN=nxA"E". 

192.  Central  Axis.  —  If  the  axis  with  reference  to  which  the 
moment  of  inertia  is  found  contains  the  centroid  of  the  given 
system  of  forces,  it  is  called  a  central  axis  of  the  system. 

In  many  cases  it  is  desired  to  find  the  moment  of  inertia  with 
respect  to  a  central  axis  whose  direction  is  known  while  the 
position  of  the  centroid  is  at  first  unknown.  It  is  to  be  noticed 
that  the  method  shown  in  Fig.  65  is  applicable  in  this  case ; 
for  the  first  part  of  the  process  is  identical  with  that  employed 
in  finding  the  centroid  of  the  system.  If,  in  Fig.  65,  the 
strings  oa,  oe,  of  the  first  funicular  polygon  be  prolonged  to 
intersect,  a  line  through  their  point  of  intersection,  parallel  to 
the  direction  of  the  forces,  will  contain  the  centroid  of  the 
system.  If  this  line  is  taken  as  the  inertia-axis,  the  points  A', 
B',  C',  D' ,  E'  are  the  points  in  which  this  axis  is  intersected  by 


MOMENTS    OF    INERTIA   OF   FORCES.  173 

the  strings  oa,  ob,  oc,  od,  oe.     No  further  modification  of  the 
process  is  necessary. 

193.  Moment  of  Inertia  Determined  from  Area  of  Funicular 
Polygon.  —  In  Fig.  65,  the  moments  of  the  given  forces  are 
represented  by  the  intercepts  A'B',  B'C',  C'D',  D'E',  each 
multiplied  by  the  pole  distance  H.  The  moment  of  inertia  of 
any  force,  as  AB,  is  equal  to  the  moment  of  a  force  represented 
by  the  corresponding  intercept  as  A'B',  supposed  to  act  in  the 
line  ab.  Now,  by  definition  (Art.  175)  the  moment  about  the 
axis  QR  of  a  force  equal  to  A'B'  acting  in  the  line  ab  is  equal 
to  double  the  area  of  the  triangle  A'  i  B'  ;  hence  the  moment  of 
the  force  fix  A'B'  is  equal  to  double  the  area  of  that  triangle 
multiplied  by//.  Similarly,  the  moment  of  a  force  HxB'C', 
acting  in  the  line  be,  is  equal  to  double  the  area  of  the  triangle 
B'  2  C'  multiplied  by  H.  Applying  the  same  reasoning  to  each 
force,  we  see  that  the  sum  of  the  moments  of  the  assumed 
forces  (HxA'B',  HxB'C',  etc.)  is  equal  to  2  //"times  the  sum 
of  the  areas  of  the  triangles  A'  I  B',  B'  2  C',  C'  3  D'  ,  £>'  ^E'. 
In  adding  these  triangles,  each  must  be  taken  with  its  proper 
sign,  corresponding  to  the  sign  of  the  moment  represented  by 
it.  Thus,  the  moments  of  A'B',  B'C',  and  D'E'  all  have  the 
same  sign,  while  the  moment  of  C'D'  has  the  opposite  sign. 
The  algebraic  sum  of  the  areas  is,  therefore, 

area  A'  i  B'  +area  B'  2  C'-area  C'  3  Z>'  +  area  D'  ^E', 


which  is  equal  to  the  area  of  the  polygon  A'  i2^^E'.  Hence 
this  area,  multiplied  by  2  H,  gives  the  moment  of  inertia  of  the 
required  system  of  forces.  ( 

It  may  sometimes  be  convenient  to  apply  this  principle  in 
determining  moments  of  inertia,  the  area  being  determined  by 
use  of  a  planimeter,  or  by  any  other  convenient  method.  It 
should  be  noticed  that  if  H  is  taken  equal  to  the  sum  of  the 
given  forces  (AE),  twice  the  area  of  the  funicular  polygon  is 
equal  to  the  square  of  the  radius  of  gyration.  If  H=\  AE,  the 


174 


GRAPHIC    STATICS. 


square  of  the  radius  of  gyration  is  equal  to  the  area  of  the 
polygon. 

194.  Determination  of  Product  of  Inertia  of  Parallel  Forces. 
—  Assume  the  points  of  application  of  the  forces  to  be  in  the 
plane  containing  the  two  axes.  If  the  moment  of  any  force 
with  respect  to  one  axis  be  found,  and  a  force  equal  in  magni- 
tude to  this  moment  be  assumed  to  act  at  the  point  of  applica- 
tion of  the  original  force,  then  the  moment  of  this  new  force 
with  respect  to  the  second  axis  is  equal  to  the  product  of 
inertia  of  the  given  force  for  the  two  axes. 

Thus,  let  ab,  be,  cd,  de  (Fig.  66)  be  the  points  of  application 
of  four  parallel  forces,  and  let  their  product  of  inertia  with 
respect  to  the  axes  QR,  ST  be  required.  Draw  ABODE,  the 


Fig.  66 

force  polygon  for  the  given  forces,  assumed  to  act  parallel  to 
QR.  Choose  a  pole  O,  the  pole  distance  being  preferably 
taken  equal  to  AE,  or  some  simple  multiple  of  AE,  and  draw 
the  funicular  polygon  as  shown,  prolonging  the  strings  to  inter- 
sect QR  in  the  points  A',  B',  C,  D\  E.  Now  assume  a  series 


r^TTlFc- 

MOMENTS    OF    INERTIA   OF   FORCES.      If  fj^foF  C. 

of  forces  equal  to  A'B',  B'C',  etc.,  each  multiplied 


at  the  points  ad,  be,  etc.,  and  determine  their  moments 
respect  to  the  axis  ST.  To  find  these  moments,  draw  lines 
through  ab,  be,  etc.,  parallel  to  ST,  and  draw  a  funicular  poly- 
gon for  the  assumed  forces  taken  to  act  in  these  lines.  The 
force  polygon  is  obtained  by  revolving  the  line  AB'C'D'E'  until 
parallel  with  ST,  and  is  the  line  A\  B\  C\  D\  E\  in  the  figure. 
The  strings  o'a',  o'e'  of  the  second  funicular  polygon  intersect 
ST  in  the  points  A"  and  E".  Hence,  calling  H'  the  second 
pole  distance,  the  product  of  inertia  of  the  given  system  is 
equal  to  HxH'xA"E". 

195.  Product-Radius.  — If  H  be  taken  equal  to  AE  (Fig.  66), 
H'xA"E"  is  equal  to  the    square  of  the  product-radius  (Art. 
189).     Hence  the  product-radius  can  be  found  by  a  construc- 
tion exactly  like  that  employed  in  finding  the  radius  of  gyra- 
tion.    Thus   (Fig.  66)  take   LM=H'  and  MN=A"E";  make 
L N  the  diameter  of  a  semicircle,  and  draw  from  M  a  line  per- 
pendicular  to   LN,   intersecting   the    semicircle   in    P.     Then 
MP2=LMx  MN=  H'xA'E";    hence    MP=c,   the    product- 
radius. 

196.  Relation  between  Moments  of  Inertia  for  Parallel  Axes. 

—  Proposition.  —  The  moment  of  inertia  of  a  system  of  par- 
allel forces  with  reference  to  any  axis  is  equal  to  its  moment  of 
inertia  with  respect  to  a  parallel  axis  through  the  centroid  of 
the  system  plus  the  moment  of  inertia  with  respect  to  the  given 
axis  of  the  resultant  applied  at  the  centroid  of  the  system. 

Let  PI,  P2,  etc.,  represent  the  forces ;  x\t  x2,  etc.,  the  dis- 
tances of  their  points  of  application  from  the  central  axis ;  and 
a  the  distance  of  this  central  axis  from  the  given  axis.  Calling 
the  required  moment  of  inertia  A,  and  the  moment  of  inertia 
with  respect  to  the  axis  through  the  center  of  gravity  A ',  we 
have 

A=Pl 


I76  GRAPHIC    STATICS. 

Now  P^  +  P^x^  —  is  the  algebraic  sum  of  the  moments  of 
the  given  forces  with  respect  to  the  axis  through  their  centroid, 
and  is  equal  to  zero  ;  and  /V'i2  +  P&s2  H  —  =  A'.  Hence 

A=A'+  (Pl  +  P,+  -}a\ 
which  proves  the  proposition. 

Radii  of  gyration.  —  Let  k  =  radius  of  gyration  of  the  sys- 
tem with  respect  to  the  given  axis,  and  k'  the  radius  of  gyra- 
tion with  respect  to  the  central  axis,  and  we  have 


Hence  the  equation  above  deduced  may  be  reduced  to  the  form 


197.  Products  of  Inertia  with  Respect  to  Parallel  Axes.  — 
Proposition.  —  The  product  of  inertia  of  a  system  of  parallel 
forces  with  reference  to  any  two  axes  is  equal  to  the  product 
of  inertia  with  reference  to  a  pair  of  central  axes  parallel  to  the 
given  axes,  plus  the  product  of  inertia  of  the  resultant  (acting 
at  the  centroid)  with  reference  to  the  given  axes. 

Let  Pv  Pz,  etc.,  be  the  magnitudes  of  the  given  forces  ;  (/b  ^), 
(A,  ^2),  etc.,  the  distances  of  their  points  of  application  from 
the  central  axes  parallel  to  the  two  given  axes  ;  (a,  b}  the  dis- 
tances of  the  centroid  of  the  system  from  the  given  axes.  Let 
A  and  A'  be  the  products  of  inertia  of  the  system  with  respect 
to  the  given  axes  and  the  parallel  central  axes  respectively. 
Then 


But  /Vi  +  /Vs+'"=°;  and  Pipi  +  P*pt-\  —  =o;  since  each  of 
these  expressions  is  the  sum  of  the  moments  of  the  given  forces 


MOMENTS    OF    INERTIA   OF   PLANE   AREAS  177 

with  respect  to  an   axis  through  the  centroid  of  the  system. 

Hence, 

A  =  A'+  (P1+Jpa+.  ..)*£, 

which  proves  the  proposition. 

If  A  =  (Pl  +  P2+-~)c2andA'=(Pl  +  P2  +  ~-y2,  we  have 


From  the  above  proposition  it  follows  that  if  the  axes  have 
such  directions  that  the  product  of  inertia  with  reference  to  the 
central  axes  is  zero,  the  product  of  inertia  with  reference  to  the 
given  axes  is  the  same  as  if  the  forces  all  acted  at  the  centroid. 
When  this  condition  is  known  to  be  satisfied,  then  for  the  pur- 
pose of  finding  the  product  of  inertia  the  system  of  forces  may 
be  replaced  by  their  resultant. 

It  follows  also,  in  the  case  when  the  product  of  inertia  for  the 
central  axes  is  .zero,  that  if  one  of  the  given  axes  coincides  with 
the  parallel  central  axis,  the  product  of  inertia  for  the  given 
axes  is  zero  ;  for  in  this  case  either  a  or  b  is  zero,  and  hence 
ab(Pi  +  Pa  +  •  •  •)  is  zero.  Therefore, 

If  the  product  of  inertia  of  a  system  is  zero  for  two  axes,  A' 
and  A",  one  of  which  (as  A')  contains  the  centroid  of  a  system, 
then  the  product  of  inertia  is  also  zero  for  A'  and  any  axis 
parallel  to  A". 

§  2.    Moments  of  Inertia  of  Plane  Areas. 

198.  Elementary  Areas  Treated  as  Forces.  —  If  any  area  be 
divided  into  small  elements,  and  a  force  be  applied  at  the 
centroid  of  each  element  numerically  equal  to  its  area,  the 
moment  of  inertia  of  this  system  of  forces  will  be  approximately 
equal  to  that  of  the  given  area.  The  approximation  will  be 
closer  the  smaller  the  elementary  areas  are  taken.  If  the  ele- 
ments be  made  smaller  and  smaller,  so  that  the  area  of  each 
approaches  zero  as  a  limit,  the  moment  of  inertia  of  the  sup- 
posed system  of  forces  approaches  as  a  limit  the  true  value  of 
the  moment  of  inertia  of  the  given  area. 


178  GRAPHIC   STATICS. 

It  is  seen,  then,  that  most  of  the  general  principles  which 
have  been  stated  regarding  moments  of  inertia  of  systems  of 
forces  are  equally  applicable  to  moments  of  inertia  of  areas. 
The  practical  application  of  these  principles,  however,  and 
especially  the  graphic  constructions  based  upon  them,  are  less 
simple  in  the  case  of  areas  than  of  systems  of  forces  such  as 
those  already  treated.  The  reason  for  this  is  that  the  system 
of  forces  which  may  be  conceived  to  replace  the  elements  of 
area  consists  of  an  infinite  number  of  infinitely  small  forces, 
with  which  the  graphic  methods  thus  far  discussed  cannot 
readily  deal.  Problems  of  this  class  are  most  easily  treated  by 
means  of  the  integral  calculus,  especially  when  the  areas  dealt 
with  are  in  the  form  of  geometrical  figures.  It  is  possible, 
however,  by  graphic  methods  to  determine  approximately  the 
moment  of  inertia  of  any  plane  area  ;  and  in  many  cases  exact 
graphic  solutions  of  such  problems  are  not  difficult.  The  proof 
of  these  methods  is  often  most  easily  effected  algebraically. 

199.  Moments  of  Inertia  of  Geometrical  Figures.  —  The  appli- 
cation of  the  integral  calculus  to  the  determination  of  moments 
of  inertia  will  not  be  here  treated.  But  the  values  of  the 
moments  of  inertia  of  some  of  the  common  geometrical  figures 
are  of  such  frequent  use  that  the  more  important  of  them  will 
be  given  for  future  reference.  The  moment  of  inertia  is  in  each 
case  taken  with  respect  to  a  central  axis,  and  will  be  repre^ 
sented  by  /,  while  the  radius  of  gyration  will  be  called  k. 

Rectangle.  —  Let  b  and  d  be  the  sides,  the  axis  being  parallel 
to  the  side  b.  Then 

jjbtf        p^ 

12    '  12 

Triangle.  —  Let  b  and  d  be  the  base  and  altitude.  Then  for 
an  axis  parallel  to  the  base, 


36 


=5^1*6" 
MOMENTS   OF   INERTIA   OF   PLANE   ARI  y^ 

For  an  axis   through  the   vertex,  bisecting  the  ^-"••^        . -«Kft« 

where  b'  is  the  projection  of  the  base  on  a  line  perpendicular  to 
the  axis. 

Circle.  —  Let^  be  the  diameter.     Then 


-.  --. 

/-            )  -        s- 

64  1  6 
For  a  central  axis  perpendicular  to  the  plane  of  the  circle, 

/=^_4.  p  =  <P 

32  '  8 

Ellipse.  —  Let  a  and  b  be  the  semi-axes.     Then  for  an  axis 
parallel  to  a, 

j_  Trafi  .  A2  _  ^ 

4     :  ~~4 
For  an  axis  parallel  to  b, 


/— 

4  4 

For  a  central  axis  perpendicular  to  the  plane  of  the  ellipse, 


Graphic  construction  for  radius  of  gyration.  —  Whenever  £* 
can  be  expressed  as  the  product  of  two  known  factors,  the  value 
of  k  can  be  found  by  the  construction  already  used  in  Art.  191. 

d^ 
Thus,  in  case  of   a  rectangle,  for  which  kz  =  —  ,  we  may  put 

p  =  d    d     Then  .f   -n  p.      6  take  LM=:d   MN=-,  the 

34  43 

construction  there  shown  will  give  MP  as  the  value  of  k.     For 

the   triangle,    the   axis   being   parallel    to   the   base,    we   have 

/£2  =  -  •  -,  and  the  same  construction  is  applicable.     For  the 
36 

axis  through  the  vertex  bisecting  the  base,  /C'2=—  •  —  • 

4      o 


!8o  GRAPHIC    STATICS. 

200.  Product  of  Inertia.  —  General  principles.  —  Products  of 
inertia  of  areas  are  determined  by  means  of  the  integral  calculus 
in  a  manner  similar  to  that  employed  for  moments  of  inertia. 
The   following    fundamental    principles    regarding   products  of 
inertia  of  geometrical  figures  will  be  found  useful : 

(1)  With  reference  to  two  rectangular  axes,  one  of  which  is 
an  axis  of  symmetry  (Art.  179),  the  product  of  inertia  is  zero. 
For  it  is  manifest  that  the  products  of  inertia  of  two  equal 
elements,   symmetrically  placed  with  reference  to  one  of   the 
axes,  are  numerically  equal  but  of  opposite  sign.      Hence,  if 
the  whole  area  can  be  made  up  of  such  pairs  of  elements,  the 
total  product  of  inertia  is  zero. 

(2)  If  the  two  axes  are  not  rectangular,  but  the  area  can  be 
divided  into  elements  such  that  for  every  element  whose  dis- 
tances from  the  axes  are  /,  q,  there  is  an  equal  element  whose 
distances  are  /,  —q,  or  —p,  q,  the  product  of  inertia  is  zero. 
This  includes  the  preceding  as  a  special  case. 

20 1.  Products  of  Inertia  of  Geometrical  Figures.  —  In  each  of 
the  following  cases  the  product  of  inertia  is  zero : 

A  triangle,  one  axis  containing  the  vertex  and  the  middle 
point  of  the  base,  the  other  being  any  line  parallel  to  the  base. 

A  parallelogram,  the  axes  being  parallel  to  the  sides  and  one 
axis  being  central.  This  includes  the  rectangle  as  a  special 
case. 

An  ellipse,  the  axes  being  parallel  to  a  pair  of  conjugate 
diameters,  and  one  axis  being  central.  This  includes,  as  a 
special  case,  that  in  which  one  axis  is  a  principal  diameter  and 
the  other  is  any  line  perpendicular  to  it ;  and  under  this  case 
falls  also  the  circle. 

202.  Approximate  Method  for  Finding  Moment  of  Inertia  of 
Any  Area.  — To  apply  the  method  of  Art.  190  to  the  determina- 
tion of  the  moment  of  inertia  of  a  plane  area,  we  should  strictly 
need  to  replace  the  area  by  an  infinite  number  of  parallel  forces, 
proportional  to  the  infinitesimal  elements  of  the  given  area,  and 


MOMENTS    OF    INERTIA   OF   PLANE   AREAS. 


181 


with  points  of  application  in  these  elements.  If,  instead,  we 
divide  the  given  figure  into  finite  portions  -whose  several  areas 
are  known,  and  assume  forces  proportional  to  those  areas  to 
act  at  their  centroids,  we  may  get  an  approximate  value  for  the 
moment  of  inertia,  which  will  be  more  nearly  correct  the  smaller 
the  elements.  This  will  be  illustrated  by  the  area  shown  in 
Fig.  67. 

Let  QR  be  the  axis  with  reference  to  which  the  moment  of 
inertia  is  to  be  found,  in  this  case  taken  as  a  central  axis. 


Kig.  67 


Divide  the  figure  into  four  rectangular  areas  as  shown,  and 
assume  forces  numerically  equal  to  these  areas  to  act  at  their 
centroids  parallel  to  QR.  The  force  polygon  is  ABCDE. 
Draw  the  funicular  polygon  corresponding  to  a  pole  O,  and  let 
the  successive  strings  intersect  QR  in  A',  B' ,  C',  D',  £'.  Take 
this  as  a  new  force  polygon,  and  with  any  convenient  pole 
distance  draw  a  second  funicular  polygon,  using  the  same  lines 
of  action.  Let  the  first  and  last  strings  intersect  QR  in  A" 
and  E".  Then  A"E"  multiplied  by  the  product  of  the  two 
pole  distances  gives  the  moment  of  inertia  of  the  four  assumed 
forces,  and  approximately  the  moment  of  inertia  of  the  given 


CS2  GRAPHIC   STATICS. 

figure.  If  the  first  pole  distance  is  taken  equal  to  AE  (as  is 
the  case  in  Fig.  67),  the  radius  of  gyration  may  be  found  by  the 
construction  of  Art.  191.  Thus  in  Fig.  67,  MP  is  the  radius 
of  gyration  as  determined  by  this  method. 

A  more  accurate  result  may  be  reached  by  dividing  the  area 
into  narrower  strips  by  lines  parallel  to  QR  ;  since  the  narrower 
such  a  strip  is,  the  more  nearly  will  the  distance  of  each  small 
element  from  the  axis  coincide  with  that  of  the  centroid  of  the 
strip.  If  the  partial  areas  are  taken  as  narrow  strips  of  equal 
width,  the  forces  may  be  taken  proportional  to  the  average 
lengths  of  the  several  strips. 

203.  Accurate  Methods.  —  If  the  given  figure  can  be  divided 
into  parts,  such  that  the  area  of  each  is  known,  and  also  its 
radius  of  gyration  with  respect  to  its  central  axis  parallel  to  the 
given  axis,  the  above  method  may  be  so  modified  as  to  give  an 
accurate  result.  Two  methods  will  be  noticed. 

(1)  When  the  axis  is  known  at  the  start.  —  Let  the  line  of 
action  of  the  force  representing  any  partial  area  be  taken  at  a 
distance  from  the  given  axis  equal  to  the  radius  of  gyration  of 
that  area  with  reference  to  the  axis.     If  this  is  done,  it  is 
evident  that  the  moment  of  inertia  of  the  system  of  forces  is 
identical  with  that  of  the  given  area.     When  the  axis  is  known, 
the  position  of  the  line  of  action  for  any  force  may  be  found  as 

follows  :  Let  QR  (Fig.  68)  be  the  given  axis, 
and  Q'R'  a  parallel  axis  through  the  centroid 
of  any  partial  area.  Draw  KL  perpendicu- 
lar to  QR,  and  lay  off  LM  equal  to  the 
radius  of  gyration  of  the  partial  area  with 
respect  to  Q'R'.  Then  KM  is  the  length 

of  the  radius  of  gyration  with  respect  to  QR.     (Art.  196.)     Take 

KN equal  to  KM  and  draw  Q"R"  through  N  parallel  to  Q'R' ; 

then   Q"R"  is  to  be  taken  as  the  line  of  action  of  the  force 

representing  the  partial  area  in  question. 

(2)  When  the  axis  is  at  first  unknown.  —  The  method  to  be 


A  (a2  +  £2)=Aa 
V 


MOMENTS   OF    INERTIA    OF   PLANE   AR 

employed  in  this  case  is  to  let  the  force  representing' 
area  act  in  a  line  through  the  centroid  of  that  area ; 
assume  the  force  representing  its  moment  to  act  in  such  a  line 
that  the  moment  of  this  second  force  shall  be  numerically 
equal  to  the  moment  of  inertia  of  the  partial  area.  This  line 
may  be  found  as  follows  :  Let  k  represent  the  radius  of  gyra- 
tion of  the  partial  area  with  respect  to  its  central  axis  parallel 
to  the  given  axis,  and  a  the  distance  between  the  two  axes. 
Then  the  moment  of  inertia  of  the  partial  area  with  respect 
to  the  given  axis  is  A  (a2  +  >£2),  if  A  represents  the  area.  But 
— \  Hence,  if  a  force  numerically  equal  to 

A  is  assumed  to  act  with  an  arm  a,  then  a  force  equal  to  its 

A2 

moment  Aa  must  act  with  an  arm  a-\ —  in  order  that  its  moment 

a 
may  equal  A  («2  +  /£2). 

The  distance  a-\ —  can  be  found  by  a  simple  construction. 
a 

Let  QR  and  Q ' R'  (Fig.  69)  be  the  given  axis  and  the  central 
axis  respectively.  Draw  KL 
perpendicular  to  QR  and  lay  off 
L M  equal  to  k.  From  M  draw 
a  line  perpendicular  to  KM,  in- 
tersecting KL  produced  at  N. 

Then  KN=a  +  k--     For>  in  the 

a 
similar  triangles  KNM,  KML,  we  have 


=       _ 
KM     KL  ' 


KL 


But  KL=a,  and  KM  =  «2  +  £2;  hence 


This  second  method  is  more  useful  than  the  first,  because  in 
applying  it  the  first  funicular  polygon  can  be  drawn  before  the 
position  of  the  inertia-axis  is  known.  Thus,  a  very  common 


1 84 


GRAPHIC    STATICS. 


case  is  that  in  which  the  moment  of  inertia  of  an  area  is  to  be 
found  for  a  central  axis,  whose  direction  is  known,  while  at  the 
outset  its  position  is  unknown  because  the  centroid  of  the  area 
is  unknown.  If  the  second  method  be  employed,  the  first 
funicular  polygon  can  be  drawn  at  once,  and  serves  to  locate 
the  required  central  axis,  as  well  as  to  determine  the  moments 
of  the  first  set  of  forces  as  soon  as  the  axis  is  known.  The 
central  axis  and  the  moment  of  inertia  with  respect  to  it  are 
thus  determined  by  a  single  construction. 

Example.  —  The  method  last  described  is  illustrated  in  Fig. 
70.  The  area  shown  consists  of  two  rectangles,  the  centroids 
of  which  are  marked  ab  and  be.  The  moment  of  inertia  is  to 
be  found  for  a  central  axis  parallel  to  the  longer  side  of  the 
rectangle  ab. 


We  draw  through  ab  and  be  lines  parallel  to  the  assumed 
direction  of  the  axis,  and  take  these  for  the  lines  of  action  of 
forces  AB,  BC,  proportional  to  the  areas  of  the  two  rectangles. 
ABC  is  the  force  polygon,  and  the  pole  distance  is  taken  equal 


MOMENTS   OF    INERTIA   OF   PLANE   AREAS.  185 

to  AC.  The  intersection  of  the  strings  oa,  oc,  determines  a 
point  of  the  required  central  axis  ac.  The  moments  of  the 
given  forces  are  proportional  to  A'B',  B'C.  We  now  have  to 
find  the  lines  of  action  for  the  forces  A'B',  B'C,  in  accordance 
with  the  method  above  described. 

Take  any  line  perpendicular  to  the  axis  ac,  as  VQ,  the  side 
of  one  of  the  rectangles.  From  R,  the  point  in  which  this  line 
intersects  the  vertical  line  through  be,  lay  off  RT  equal  to  the 
central  radius  of  gyration  of  the  rectangle  whose  centroid  is 
be.  To  find  this  central  radius  of  gyration,  we  know  that  its 

IS* 


value  is  -y — -  (Art.  199),  where  d  is  the  length  of  the  side  of 
the  rectangle  perpendicular  to  the  axis.  Hence  we  take 

QR  =  -,  RS=-,  and   make  QS  the  diameter  of  a  semicircle, 

3  4 

intersecting  the  vertical  line  be  in  T;  then  RT  is  the  required 
radius  of  gyration  of  the  rectangle  with  respect  to  a  central 
axis.  Now  draw  from  T  a  line  perpendicular  to  VT,  intersect- 
ing VQ  in  U;  then  the  line  b'c'  drawn  through  £7  parallel  to  the 
given  axis  is  the  line  of  action  of  the  force  B'C'. 

By  a  similar  construction  applied  to  the  other  rectangle,  a'b' 
is  located  as  the  line  of  action  of  the  force  A'B'.  The  second 
funicular  polygon  is  now  drawn,  and  the  points  A",  C"  are 
found  by  the  intersection  of  the  strings  o'a',  o'c'  with  the  axis. 
Hence  the  moment  of  inertia  of  the  given  area  is  equal  to 
A"C"xHxH'.  In  the  figure  H  is  made  equal  to  AC,  hence 
the  radius  of  gyration  can  be  determined  by  the  usual  con- 
struction, and  its  length  is  found  to  be  MP. 

204.  Moment  of  Inertia  of  Area  Determined  from  Area  of 
Funicular  Polygon.  —  The  method  given  in  Art.  193  for  finding 
the  moment  of  inertia  of  a  system  of  forces  by  means  of  the 
area  of  the  funicular  polygon  may  be  applied  with  approximate 
results  to  the  case  of  a  plane  figure.  If  the  forces  are  taken 
as  acting  at  the  centroids  of  the  areas  they  represent,  then  to 


1 86  GRAPHIC   STATICS. 

get  good  results  these  partial  areas  should  be  taken  as  narrow 
strips  between  lines  parallel  to  the  axis. 

If  the  lines  of  action  are  determined  as  in  the  first  method 
of  the  preceding  article,  then  the  area  enclosed  by  the  funic- 
ular polygon  and  the  axis  represents  accurately  the  required 
moment  of  inertia. 

205.  Product  of  Inertia  Determined  Graphically.  —  To  deter- 
mine the  product  of  inertia  of  any  area,  let  it  be  divided  into 
small  known  parts,  and  let  parallel  forces  numerically  equal  to 
the  partial  areas  be  assumed  to  act  at  the  centroids  of  these 
parts.  The  product  of  inertia  of  these  forces  may  then  be 
found  as  in  Art.  194,  and  its  value  will  represent  approximately 
the  product  of  inertia  of  the  given  area. 

If  the  partial  areas  can  be  so  taken  that  the  product  of 
inertia  of  each  with  reference  to  axes  through  its  centroid  par- 
allel to  the  given  axes  is  zero,  the  method  here  given  is  exact. 
(Art.  197.) 

If  the  partial  areas  are  taken  as  narrow  strips  parallel  to  one 
of  the  axes,  the  condition  just  mentioned  will  be  nearly  fulfilled  ; 
for  the  product  of  inertia  of  each  strip  for  a  pair  of  axes 
through  its  centroid,  one  of  which  is  parallel  to  its  length,  will 
be  very  small. 


CHAPTER   X.      CURVES   OF   INERTIA. 

§    I.    General  Principles. 

206.  Relation  between  Moments  of  Inertia  for  Different  Axes 
through  the  Same  Point.  — The  moment  of  inertia  with  respect 
to  any  axis  through  a  given  point  can  be  expressed  in  terms  of 
the  moments  and  product  of  inertia  for  any  two  axes  through 
that  point.  It  is  necessary  here  to  use  algebraic  methods,  but 
the  results  reached  form  the  basis  of  graphic  constructions. 

Let  OX,  OY  (¥\g.  71)  be  the  two  given  axes;  6,  the  angle 
included  between  them  ;  Pu  P2,  etc.,  the  forces  of  the  system  ;  *', 
y,  the  coordinates  of  the  point  / 

of  application  of  any  force  P, 
referred  to  the  axes  OX,  OY; 
p,  q,  the  perpendicular  distances 
of  the  same  point  from  (9Kand 
OX  respectively,  so  that  i 

<^/  x 

/=*'sm0,    g=y'sm0.  /  Fis- 71 

Let  a',  b',  and  c'  be  quantities  denned  by  the  equations 


187 


1  88  GRAPHIC   STATICS. 

Then 


and  a'  sin  0,  b'  sin  0,  c'  sin  6  are  respectively  the  radius  of 
gyration  with  respect  to  O  Y,  the  radius  of  gyration  with  respect 
to  OX,  and  the  product-radius  (Art.  189)  with  respect  to  OX 
and  OY. 

The  moment  of  inertia  (/)  and  radius  of  gyration  (k)  of  the 
system  for  the  axis  OM,  making  an  angle  <f>  with  OX,  may  now 
be  computed  as  follows  : 

Let  s  =  perpendicular  distance  of  the  point  of  application  of 
any  force  P  from  OM.  Then  from  the  geometry  of  the  figure 
it  is  seen  that 

s=y'  sin  (6  —  <£)  —x1  sin  <f>. 
Hence 


sin  (0  -  </>)  sin 
+  2/V2  sin2  <£; 

sin2  (e-$)-2c'*2P  .  sin  (0-0)  sin  0 


the  factors  involving  0  and  0  being  constant  for  all  terms  of 
the  summation.     Hence 

sin20   .   .    (i) 

From  these  equations  /  and  k  may  be  computed  if  a',  b',  and 
cr  are  known  ;  that  is,  if  the  moments  and  product  of  inertia  for 
the  two  axes  OX  and  O  Y  are  known. 

Special  case.  —  If  #  =  90°,  the  equation  (i)  becomes 

/^  =  ^'2cos20-2^2sincos+fl'2sin2<>  ........  (2) 


GENERAL   PRINCIPLES.  jSQ 

207.  Products  of  Inertia  for  Different  Axes  through  the  Ori- 
gin. —  The  product  of  inertia  with  respect  to  OM  and  OX  may 
be  found  as  follows  : 

Let    A  =  the  required  product  of  inertia  ;  then 

A  =  ^Pqs  =  ^Py'  sin0  [/  sin  (0  —  0)—  *'  sin0] 
=  2P  [/2  sin  0  sin  (0-0)  -*'/  sin  0  sin  0] 
=  [£'2  sin  0  sin  (0  -  0)  -  c  '2  sin  0  sin  0]  2/>. 
Let  //  =  product-radius  for  axes  OM  and  OX;  then 

/P  =     -  =  b'*  sin0  sin  (0-0)-^'2  sin  0  sin  0. 


Special  case.  —  The  axis  0J/  may  be  so  chosen  that  A=o. 
This  will  be  the  case  if 


208.  Inertia  Curve.  —  If  on  OM  (Fig.  71)  a  point  M  be  taken 
such  that  the  length  OM  depends  in  some  given  way  upon  the 
value  of  k,  and  if  similar  points  be  located  for  all  possible  direc- 
tions of  OM,  the  locus  of  such  points  will  be  a  curve  which  is 
called  a  curve  of  inertia  of  the  system  for  the  center  O. 

The  form  of  the  curve  will  depend  upon  the  assumed  law 
connecting  OM  with  k. 

209.  Ellipse  or  Hyperbola  of  Inertia.  —  The  simplest  curve  is 
obtained   by  assuming  OM  to  be  inversely  proportional  to  k. 
Let  OM=r,  and   take  r*=d*^e,    where    d*    is    a    positive 

K 

quantity,  so  that  d  always  represents  a  real  length,  positive  or 
negative. 

Equation  (i)  of  Art.  206  then  becomes 
d*  sin2  0  =  b'  V  sin2  (0  -0)  -2  ^'V2  sin  (0-0)  sin  0+«'V  sin2  0, 
which  is  the  polar  equation  of  the  inertia-curve,  r  and  0  being 
the  variable  coordinates.     Let  x,  y  be  coordinates  of  the  point 
M  referred  to  the  axes  OX,  O  Y.     Then 


sin0     sin  (0  —  0)     sin  0 


190  GRAPHIC   STATICS. 

whence  * 

r*  sin2 


sin20          ' 

2  sin(#  —  (ft)  sin(ft 


sin20 
and  the  equation  becomes 

<P  =  6'W-2c'2xy  +  afy   ...........   (3) 

The  form  of  this  equation  shows  that  it  represents  a  conic 
section  whose  center  is  at  the  origin  of  coordinates  O.  This 
conic  may  be  either  an  ellipse  or  a  hyperbola. 

The  equation  will  be  discussed  more  fully  in  a  later  article. 
One  fact  may,  however,  be  here  noticed. 

If  the  moment  of  inertia  7  is  positive  for  all  positions  of  the 
axis,  the  radius  of  gyration  k  will  be  real,  whatever  the  value 
of  (ft.  But  r,  the  radius  vector  of  the  curve,  will  be  real  when  k 
is  real;  hence  in  this  case  the  curve  is  an  ellipse.  This  is 
always  the  case  if  the  given  forces  have  all  the  same  sign. 

If  the  forces  have  not  all  the  same  sign,  it  is  possible  that 
the  value  of  7  may  have  different  signs  for  different  directions 
of  the  axis.  If  this  is  so,  certain  values  of  (ft  make  k  (and 
therefore  r)  imaginary.  In  this  case  the  curve  is  a  hyperbola. 

The  most  important  case  is  that  in  which  the  given  forces 
have  all  the  same  sign  which  may  be  taken  as  plus,  so  that  the 
moment  of  inertia  is  always  positive,  and  the  curve  an  ellipse  ; 
and  to  this  case  the  discussion  will  be  confined. 

§  2.   Inertia-Ellipses  for  Systems  of  Forces. 

210.  Properties  of  the  Ellipse.  —  In  discussing  ellipses  of  in- 
ertia use  will  be  made  of  certain  general  properties  of  the 
ellipse,  which,  for  convenience  of  reference,  will  be  here  sum- 
marized. For  the  proof  of  the  propositions  stated  the  reader 
is  referred  to  works  on  the  conic  sections. 


INERTIA-ELLIPSES   FOR   SYSTEMS   OF   FOR 
(i)  The  equation 


represents  an  ellipse  if  IP—AC  is  negative;  a  hyperbola  if 
B2  —  AC  is  positive.  (Salmon's  Conic  Sections,  p.  140.)  The 
coordinate  axes  may  be  either  rectangular  or  oblique. 

(2)  Two  diameters  of  an  ellipse  are  said  to  be  conjugate  to 
each  other  if  each  bisects  all  chords  parallel  to  the  other.  If 
the  axes  of  coordinates  coincide  with  a  pair  of  conjugate 
diameters,  the  lengths  of  which  are  2  a'  and  2  b',  the  equation 
of  the  curve  is 


A  particular  case  of  this  equation  is  that  in  which  the  coordi- 
nate axes  are  rectangular,  being  the  principal  axes  of  the  curve  ; 
in  which  case  we  may  write  a  and  b  instead  of  a'  and  b'. 

(3)  In  an  ellipse,  the  product  of  any  semi-diameter  and  the 
perpendicular  from  the  center  on  the  tangent  parallel  to  that 
semi-diameter  is  constant  and  equal  to  ab.     That  is,  if  r  is  any 
radius  vector  of  the  curve  drawn  from  the  center,  and  /  the 
length   of  the   perpendicular  from  the  center  to  the  parallel 
tangent,  we  have 

pr=ab 
where  a  and  b  are  the  principal  semi-axes  of  the  curve. 

(4)  Let  a'  and  b'  be  conjugate  semi-diameters.     Then  each  is 
parallel  to  the  tangent  at  the  extremity  of  the  other.     Hence 
the  length  of  the  perpendicular  from  the  center  to  the  tangent 
parallel  to  a'  is  b'  sin  6,  where  B  is  the  angle  included  between 
a'  and  b'.     Therefore  from  the  preceding  paragraph, 

a'b's'm  e  =  ab. 

(5)  An  ellipse  can  be  constructed,  when  a  pair  of  conjugate 
diameters  is  known,  as  follows  : 


IQ2  GRAPHIC    STATICS. 

Let  A  A',  BB'  (Fig.  72),  be  the  conjugate  diameters,  O  being 
the  center  of  the  ellipse.  Complete  the  parallelogram  OBCA. 

Divide   OA  and   CA   into  parts 

-  -r^i  -  ~^7      proportional  to  each  other,  be- 
/    ^'"/^y        ginning  at  O  and   C.     Through 
^  _  I*'''  ^s'''  /  the   points    of    division    of    OA 

/''        /  draw   lines    radiating    from   B'  , 

_  /  and  through  the  points  of  divi- 

B'  sion  of  CA  draw  lines  radiating 

from   B.     The   points   of  inter- 

section of  the  corresponding  lines  in  the  two  sets  are  points  of 
the  ellipse.  In  a  similar  way,  the  other  three  quadrants  may 
be  drawn.  (The  location  of  one  point  is  shown  in  the  figure.) 

A  convenient  way  to  locate  the  corresponding  points  of 
division  on  OA  and  CA  is  to  cut  these  lines  by  lines  parallel  to 
the  diagonal  OC. 

211.  Discussion  of  Equation  of  Inertia-Curve.  —  We  will  now 
examine  the  equation  of  the  inertia-curve, 


with  reference  to  the  properties  of  the  ellipse  above  enumerated. 

(i)  If  a'2d'2—c'*  is  positive,  the  equation  denotes  an  ellipse. 
This  cannot  be  the  case  if  a'2  and  bn  have  opposite  signs.  But 
from  the  definitions  of  a'z  and  b'2  (Art.  206)  it  is  seen  that  their 
signs  are  the  same  as  those  of  the  moments  of  inertia  for  Fand 
X  axes  respectively.  Hence,  if  there  are  any  two  axes  through 
the  assumed  center  for  which  the  moments  of  inertia  have 
opposite  signs,  the  inertia-curve  is  a  hyperbola. 

If  the  moment  of  inertia  has  the  same  sign  for  all  axes 
through  the  assumed  center,  the  curve  is  an  ellipse.  For,  since 
c'z  sin2#  =  ^  (Art.  206),  c'  may  be  made  zero  by  choosing  the 

axes  so  that  the  product  of  inertia  with  respect  to  them  is 
zero  ;  and  if  c'  is  zero,  and  a'2  and  bn  have  the  same  sign,  the 
quantity  anbn—c*  is  positive. 


INERTIA-ELLIPSES   FOR   SYSTEMS   OF   FORCES. 


193 


This  agrees  with  the  conclusion  stated  in  Art.  209. 
We  shall  here  deal  only  with  ellipses  of  inertia. 

(2)  If  c'  =  o,  the  coordinate  axes  are  conjugate  axes  of  the 
curve.     But   the   condition   c'  =  o   means   that  the   product   of 
inertia  for  the  two  axes  is  zero.     Hence  any  two  axes  for  which 
the  product  of  inertia  is  zero  are  conjugate  axes  of  the  inertia- 
curve.     (This  is  true   whether  the   curve   is   an   ellipse   or   a 
hyperbola.) 

(3)  By  the  law  of  formation  of  the  inertia-conic  (Art.  209), 
the  length  of  the  radius  vector  lying  in  any  line  is  inversely 
proportional  to  the  radius  of  gyration  with  respect  to  that  line. 
But  by  (3)  of  the  last  article,  the  perpendicular  from  the  center 
on  the  tangent  parallel  to  any  radius  vector  is  inversely  propor- 
tional to  the  length  of  that  radius  vector.     Hence  the  perpen- 
dicular distance  between  any  diameter  and  the  parallel  tangent 
is  directly  proportional  to  the  radius  of  gyration  with  respect  to 
that   diameter.     The   curve   may  be  so  constructed   that   the 
length  of  this  perpendicular  is  equal  to  the  radius  of  gyration, 
as  follows  : 

From  Art.  209,  we  have 


r 

and  from  (3)  and  (4)  of  the  last  article  we  have 
ab     a'b'  sin  6 

/=7=  ^7"—' 

if  a'  and  V  are  conjugate  semi-diameters.     Now  take 


or  =  a, 

and  we  have  k—p, 

and  the  equation  of  the  curve  becomes  (since  c'=o  when  the 

axes  are  conjugate) 


194  GRAPHIC   STATICS. 

If  the  equation  be  written  in  this  form,  a'  and  b1  having  the 
meanings  assigned  in  Art.  206,  the  radius  of  gyration  about  any 
axis  through  the  center  of  the  ellipse  is  equal  to  the  perpendicular 
distance  between  the  axis  and  the  parallel  tangent  to  the  ellipse. 

Hereafter  we  shall  mean  by  inertia-ellipse  the  curve  obtained 
by  taking  dz=a'b'  as  above  described,  so  that  the  radius  of  gyra- 
tion for  any  axis  can  be  found  by  direct  measurement  when  a 
parallel  tangent  to  the  ellipse  is  known. 

212.  To  Determine  Tangents  to  the  Inertia-Ellipse  for  Any 
Center.  —  Let   the  radius   of  gyration   (k)    be   found   for   any 
assumed  axis  through  the  given  center  by  one  of  the  methods 
already  described  (Arts.  202  and  203).     Then   two   lines  par- 
allel to  the  axis  and  distant  k  from  it,  on  opposite  sides,  will  be 
tangents  to  the  inertia-ellipse. 

213.  To  Construct  the  Inertia-Ellipse,  a  Pair  of  Conjugate 
Axes  Being  Known  in  Position.  —  If  the  positions  of  two  axes 
conjugate  to  each  other  can  be  found,  the  ellipse  can  be  drawn 
by  the  following  method : 

Determine  the  radius  of  gyration  for  each  of  the  two  axes 
and  draw  the  corresponding  tangents  as  in  the  preceding 
article ;  then  proceed  as  follows : 

Let  XX ',  YY'  (Fig.  73)  be  the  given  axes,  and  let  the  four 

tangents  determined  as  above  form  the  parallelogram  PQRS. 

/Y  Let  A,  A',  B,  B'  be  the  points 

P J^- yQ.      in   which    these    tangents    in- 

/  /  /x       tersect   the   axes   XX',    YY'. 

X/  ~TQ  /A Then,  since  each  diameter  is 

/  / /  parallel  to  the  tangents  at  the 

S          &/  R  extremities   of   the    conjugate 

/Y'  Fis.va  diameter,  A,  A',  B,  B'  are  the 

extremities    of   the   diameters   lying  in  the  given   axes.     The 

ellipse    can    now    be    constructed   as   explained   in   Art.    210 

(Fig,  72). 

This    method  of    constructing    the    inertia-ellipse   is   useful 


INERTIA-CURVES   FOR   PLANE   AR 

whenever  the  given  system  has  a  pair  of  conjug^ 
can  be  located  by  inspection. 

214.  Central  Ellipse.  —  It   is   evident   that    an   inertia-curve 
can  be  found  with  its    center   at   any  assumed   point.     That 
ellipse  whose  center  is  the  centroid  of  the  given  system  is  called 
the  central  ellipse  for  the  system. 

Since  the  central  ellipse  gives  at  once  the  radius  of  gyration 
for  every  axis  through  the  centroid  of  the  system,  it  enables  us 
to  determine  readily  the  radius  of  gyration  for  any  axis  what- 
ever, by  means  of  the  known  relation  between  radii  of  gyration 
for  parallel  axes.  (Art.  196.) 

§  3.    Inertia-Curves  for  Plane  Areas. 

215.  General  Principles.  —  The    principles   deduced   in   the 
treatment  of  inertia-curves  for  systems  of  forces  are  all  true  for 
the  case  of  plane  areas.     But  special  difficulties  arise  in  dealing 
with  areas,  because  of  the  fact  that  the  system  of  forces  equiva- 
lent to  any  area  consists  of  an  infinite  number  of  forces.     The 
principles  already  developed  are,  however,  sufficient  to  deal  at 
least  approximately  with  all  areas,  and  accurately  with  many 
cases. 

216.  Inertia-Curve  an  Ellipse.  —  Since  the  forces  conceived 
to  replace  the  elements  of  area  (Art.   198)  have  all  the  same 
sign,  the  value  of  kz  is  always  positive,  and  the  inertia-curve  is 
always  an  ellipse.     (Arts.  209  and  211.) 

217.  Cases  Admitting  Simple  Treatment.  —  Whenever  a  pair 
of  conjugate  diameters  can  be  located,  and  the  radius  of  gyra- 
tion determined  for  each,  the  inertia-ellipse   can    be   at   once 
drawn  as  in  Art.  213.     This  will  be  the  case  whenever  it  is 
possible  to  locate  readily  a  pair  of  axes  for  which  the  product  of 
inertia  is  zero. 

(i)  If  there  is  an  axis  of  symmetry,  this  and  any  line  perpen- 
dicular to  it  are  a  pair  of  conjugate  axes  (and  in  fact  the  principal 
axes)  of  the  inertia-ellipse  whose  center  is  at  their  intersection. 
(Art.  200.) 


196  GRAPHIC    STATICS. 

(2)  If  two  axes  can  be  located  in  such  a  way  that  for  every 
element  of  area  whose  distances  from  the  axes  are/,  q,  there  is 
an  equal  element  whose  distances  are  /,  —  q,  or  —p,  q,  the 
product  of  inertia  is  zero  for  the  two  axes,  and  these  are  there- 
fore a  pair  of  conjugate  axes  of  the  inertia-ellipse  whose  center 
is  at  their  intersection.  This  of  course  includes  the  case  when 
there  is  an  axis  of  symmetry.  (Art.  200.) 

When  a  pair  of  conjugate  axes  is  known,  the  radius  of  gyration 
is  to  be  found  by  one  of  the  methods  of  Art.  202  or  Art.  203  ; 
the  ellipse  can  then  be  drawn  exactly  as  explained  in  Art.  213. 

If  a  pair  of  conjugate  axes  cannot  be  located  by  inspection, 
the  inertia-ellipse  cannot  be  so  readily  constructed.  Such  cases 
will  not  be  here  treated. 

As  examples  of  areas,  in  which  the  principal  axes  of  the 
inertia-curve  can  be  located  by  inspection,  may  be  mentioned 
the  cross-section  of  the  I-beam,  the  deck-beam,  the  channel-bar, 
and  other  shapes  of  structural  iron  and  steel. 

Many  geometrical  figures  possess  axes  of  symmetry.  In 
others  a  pair  of  conjugate  axes  can  be  located  by  principle  (2). 
Some  of  these  will  be  discussed  in  the  next  article. 

Example.  —  Draw  the  central  ellipse  for  the  deck-beam  sec- 
tion shown  in  Fig.  74. 

[SUGGESTIONS.  —  Since  there  is  an  axis  of  symmetry,  this 
contains  one  of  the  principal  axes  of  the  ellipse.  The  other 
can  be  drawn  as  soon  as  the  centroid  of  the  section  is 
known.  Find  the  radius  of  gyration  for  each  axis  by 
Art.  202,  and  then  construct  the  ellipse  as  explained  in 
Art.  213.] 

2 1 8.    Central  Ellipses  for  Geometrical  Fig- 
ures. —  In  many  of  the  simple  geometrical 
figures,   not    only  can  a   pair  of   conjugate 
axes  be  located  by  inspection,  but  the  radius 
of  gyration  for  each  of  these  axes  can  be  found  by  a  simple 
construction,  so  that  the  central  ellipse  can  be  readily  drawn. 
Some  of  these  cases  will  be  here  summarized. 


INERTIA-CURVES   FOR   PLANE   AREAS. 


197 


(i)  Parallelogram.  —  Let  ABCD  (Fig.  75)  be  the  parallelo- 
gram ;  then  XX ',  YY',  drawn  through  the  centroid  parallel  to 
the  sides,  are  a  pair  of  . 

conjugate  axes  of  the  cen-  A  / 

tral  ellipse.  Let  AB  =  b, 
BC=d,  and  let  //=  the 
perpendicular  distance  be- 
tween AB  and  DC.  The 
moment  of  inertia  of  the 
parallelogram  with  re- 
spect to  the  axis  XX'  is 
equal  to  the  moment  of  inertia  of  a  rectangle  of  sides  b  and 

Hence  /£2,  the  square  of  the  radius  of  gyration  for  this  axis,  is 
7/2 
— •     The  length  of  k  can  be  found  by  the  construction  used  in 

case  of  the  rectangles  in  Fig.  70.     The  following  modification 
of  the  method  is,  however,  more  convenient  : 

Make  EF=\BC,  EG=\BC,  and  draw  a  semicircle  with  FG 
as  a  diameter.  From  E  draw  a  line  perpendicular  to  BC,  inter- 
secting the  semicircle  at  /.  Lay  off  EH=EI;  then  a  line 
through  H  parallel  to  XX '  is  a  tangent  to  the  central  ellipse. 
For  by  construction, 

EH   El    ^ 

And  since  the  projection  of  BC  on  a  line  perpendicular  to  XX is 

equal  to  A,  the  projection  of  EH  on  the  same  line  is  equal  to  — -f— , 

VI2 
that  is  to  k. 

The  tangent  parallel  to  the  side  BC  may  be  found  in  a  simi- 
lar way.  It  may,  however,  be  located  more  simply  as  follows  : 
It  is  evident  that  the  distance  between  YY'  and  a  tangent  par- 
allel to  it,  measured  along  AB,  bears  the  same  ratio  to  AB  that 
EH  does  to  BC.  Hence,  the  parallelogram  formed  by  the  four 
tangents,  two  parallel  to  XX'  and  two  parallel  to  YY',  is  simi- 
lar to  the  parallelogram  ABCD. 


I98 


GRAPHIC   STATICS. 


Fig.  75  shows  this  parallelogram  and  also  the  ellipse. 

(2)  Triangle.  —  Let  ABC  (Fig.  76)  be  the  triangle;  b  = 
length  of  the  base  BC;  b'  = 
length  of  projection  of  BC  on  a 
line  perpendicular  to  AD ;  d  =  al- 
titude measured  perpendicular  to 
j  BC.  AD  and  a  line  through  the 
centroid  parallel  to  BC  are  a  pair 
of  conjugate  axes  of  the  central 
ellipse  (Art.  217).  From  Art. 
199,  the  radius  of  gyration  for  a 
central  axis  parallel  to  BC  is 

be  this  axis,  and  H  the  point  in  which 

it  intersects  A C.  Then  HC=  \AC.  Take  HK=  \ A C=  \ HC, 
and  make  KC  the  diameter  of  a  semicircle.  From  H  draw  HI 
perpendicular  to  AC,  intersecting  the  semicircle  at  /.  Make 
HL=HI\  then  the  line  through  L  parallel  to  XX'  is  a  tangent 
to  the  central  ellipse.  For  the  radius  of  gyration  with  respect 
to  XX'  is  to  HL  as  the  altitude  d  is  to  AC. 


Again,  for  the  axis  AD,  the  radius  of  gyration  is  -y —  (Art. 

199).  Make  DE=\  BC  and  DF=\  BC,  and  take  EF  as  a 
diameter  of  a  semicircle.  From  D  draw  a  line  perpendicular  to 
BC,  intersecting  the  semicircle  in  M;  and  make  DG=DM\  then 
a  line  from  £  parallel  to  AD  is  a  tangent  to  the  central  ellipse. 
The  figure  shows  the  parallelogram  formed  by  the  two  tan- 
gents parallel  to  XX'  and  the  two  parallel  to  AD,  and  also  the 
central  ellipse. 

(3)  Ellipse.  —  From  Art.  199,  the  radii  of  gyration  of  an 
ellipse  with  respect  to  the  two  principal  diameters  are  |« 
and  ^  b.  Hence  the  central  ellipse  of  inertia  is  similar  to  the 
given  ellipse,  its  semi-axes  being  %  a  and  ^  b.  A  special  case 
of  this  is  a  circle,  for  which  the  central  curve  is  a  circle  whose 
radius  is  half  that  of  the  given  circle. 


INERTIA-CURVES   FOR   PLANE   AREAS. 


199 


X' 


(4)    Semicircle.  —  Let    ABC   (Fig.  77)  be  the  semicircle,   O 
being   the   centroid.      (The   point   O  may  be   located   by  the 
method  described  in  Art.  184.) 
From   symmetry  it  is  evident 
that  the  principal  axes  of  the 
central    ellipse    are   XX!   and 
yy,    drawn    through    O,    re- 
spectively   parallel    and     per- 
pendicular to  AB. 

With  respect  to  the  axis 
yy,  the  radius  of  gyration 
is  evidently  -|-  r,  the  same  as 

for  the  whole  circle.     Hence  two  lines  parallel  to  yy  and  dis- 
tant ^  r  from  it  are  tangents  to  the  central  ellipse. 

Again,  the  radius  of  gyration  of  the  semicircle  with  respect 
to  AB  as  an  axis  is  also  ^  r,  the  same  as  for  the  whole  circle. 
To  find  it  for  the  axis  XX',  with  D  as  a  center,  and  radius  \  r, 
draw  an  arc  intersecting  XX'  at  F;  then  IJF*  =  DF* ~  (?& . 
But  DFis  equal  to  the  radius  of  gyration  with  respect  to  AB, 
and  ODis  the  distance  between  XX'  and  AB ;  hence  (Art.  196) 
OF  is  equal  to  the  radius  of  gyration  with  respect  to  XX'. 
Hence  if  two  lines  are  drawn  parallel  to  XX',  each  at  a  distance 
from  it  equal  to  OF,  they  will  be  tangents  to  the  central  ellipse. 
The  ellipse  can  now  be  drawn  in  the  usual  manner. 

219.  Summary  of  Results.  —  By  the  principles  and  methods 
developed  in  the  present  chapter,  inertia-curves  can  be  drawn 
for  all  the  simpler  cases  that  may  arise ;  namely,  whenever  a 
pair  of  conjugate  axes  can  be  located  by  inspection.  This  will 
be  the  case  whenever  the  product  of  inertia  can  be  seen  to  be 
zero  for  any  pair  of  axes  ;  and  it  includes  every  case  of  an  area 
possessing  an  axis  of  symmetry. 

It  is  believed  that  this  chapter  contains  as  complete  a  discus- 
sion as  is  needed  by  the  student  of  engineering.  Those  who 
desire  to  pursue  the  subject  further  may  consult  other  works. 


Fig.  34= 


(B) 

Scale,  1  inch=1800  Ibs. 


PLATE  I. 


DEF 

Scale,  1  inch=6,ooo  Ibs. 
(CO 


FG 


Fig.  42 


Scale:  I  in.=  4.OOO  Ibs. 


PLATE    II. 


DEFG 


U.  OF  C. 


§11?  I 

I  "  I"  I  -1   I 


(S) 


Linear  Scale,  1  in.  =  20  ft. 
Force  Scale,  1  in.  =  80,000  Ibs. 


PLATE  III. 


I]  S  !    ill  II 

I    I    1 5   I    15    |        7.1       I     j.8   I      5  7     I    4.8    \l  0 


oo     o    OOOO   oo  OQ  t 


15()0  l\js.  per  foot 


JB, 


G  „-• 


3SOC1A 


B, 


(A) 


3'        4'       5'          6 


•.„      F 


G    "- 


7'       8'        9'       \"  X 


B\ 


U.  OF  C. 


PLA' 


iear  Scale.  1  in.- 20  ft. 
b?e  Scale,  1  in.- SO, 000  Ibs. 


PLATE  VI. 


in.-  20  ft. 
n.=  SO, 000  Ibs, 


C,  D 


A.*-- 


PLATE  VII. 


Linear  Scale,  1  in.-  CO  ft. 
Force  Scale,!  in.-  120,000  Ibs 


Bi 


(A) 


PLATE  VIII. 


Fig.  49 


Fig.  50 


THE  ELEMENTS  OF 


EDWARD  L.   NICHOLS,   B.S.,   Ph.D., 

Professor  of  Physics  in  Cornell  University, 


WILLIAM  S.   FRANKLIN,  M.S., 

Professor  of  Physics  and  Electrical  Engineering  at  Lehigh  University. 
WITH    NUMEROUS    ILLUSTRATIONS. 

PART  I.     IN  THREE  VOLUMES: 

Vol.  I.  Mechanics  and  Heat  .  .  Price  $1.50  net. 
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It  has  been  written  with  a  view  to  providing  a  text-book  which  shall  correspond  with 
the  increasing  strength  of  the  mathematical  teaching  in  our  university  classes.  In  most  of 
the  existing  text-books  it  appears  to  have  been  assumed  that  the  student  possesses  so 
scanty  a  mathematical  knowledge  that  he  cannot  understand  the  natural  language  of 
physics,  i.e.,  the  language  of  the  calculus.  Some  authors,  on  the  other  hand,  have  assumed 
a  degree  of  mathematical  training  such  that  their  work  is  unreadable  for  nearly  all  under- 
graduates. 

The  present  writers  having  had  occasion  to  teach  large  class  -,  the  members  of  which 
were  acquainted  with  the  elementary  principles  of  the  calculus,  have  sorely  felt  the  need  of 
a  text-book  adapted  to  their  students.  The  present  work  is  an  attempt  on  their  part  to 
supply  this  want.  It  is  believed  that  in  very  many  institutions  a  similar  condition  of  affairs 
exists,  and  that  there  is  a  demand  fir  a  work  of  a  grade  intermediate  between  that  of  the 
existing  elementary  texts  and  the  advanced  manuals  of  physics. 

No  attempt  has  been  made  in  this  work  to  produce  a  complete  manual  or  compendium 
of  experimental  physics.  The  book  is  planned  to  be  used  in  connection  with  illustrated 
lectures,  in  the  course  of  which  the  phenomena  are  demonstrated  and  described.  The 
authors  have  accordingly  confined  themselves  to  a  statement  of  principles,  leaving  the 
lecturer  to  bring  to  notice  the  phenomena  based  upon  them.  In  stating  these  principles, 
free  use  has  been  made  of  the  calculus,  but  no  demand  has  been  made  upon  the  student 
beyond  that  supplied  by  the  ordinary  elementary  college  courses  on  this  subject. 

Certain  parts  of  physics  contain  real  and  unavoidable  difficulties.  These  have  not  been 
slurred  over,  nor  have  those  portions  of  the  subject  which  contain  them  been  omitted.  It 
has  been  thought  more  serviceable  to  the  student  and  to  the  teacher  who  may  have  occa- 
sion to  use  the  book  to  face  such  difficulties  frankly,  reducing  the  statements  involving 
them  to  the  simplest  form  which  is  compatible  with  accuracy. 

In  a  word,  the  Elements  of  Physics  is  a  book  which  has  been  written  for  use  in  such 
institutions  as  give  their  undergraduates  a  reasonably  good  mathematical  training.  It  is 
intended  for  teachers  who  desire  to  treat  their  subject  as  an  exact  science,  and  who  are 
prepared  to  supplement  the  brief  subject-matter  of  the  text  by  demonstration,  illustration, 
and  discussion  drawn  from  the  fund  of  their  own  knowledge. 


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A  Laboratory  flanual 

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Physics  and  Applied  Electricity. 

ARRANGED   AND   EDITED   BY 

EDWARD    L.    NICHOLS, 

Professor  of  Physics  in   Cornell  University. 
IN    TWO    VOLUMES. 

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assumed  that  the  student  possesses  some  knowledge  of  analytical  geometry  and  of  the  cal- 
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ligence of  the  practicant. 

A  large  proportion  of  the  students  for  whom  primarily  this  Manual  is  intended,  are  pre- 
paring to  become  engineers,  and  especial  attention  has  been  devoted  to  the  needs  of  that 
class  of  readers.  In  Vol.  II.,  especially,  a  considerable  amount  of  work  in  applied  elec- 
tricity, in  photometry,  and  in  heat  has  been  introduced. 

COMMENTS. 

"The  work  as  a  whole  cannot  be  too  highly  commended.  Its  brief  outlines  of  the 
various  experiments  are  very  satisfactory,  its  descriptions  of  apparatus  are  excellent;  its 
numerous  suggestions  are  calculated  to  develop  the  thinking  and  reasoning  powers  of  the 
student.  The  diagrams  are  carefully  prepared,  and  its  frequent  citations  of  original 
sources  of  information  are  of  the  greatest  value."  —  Street  Railway  Journal. 

"  The  work  is  clearly  and  concisely  written,  the  fact  that  it  is  edited  by  Professor  Nichols 
being  a  sufficient  guarantee  of  merit."  —  Electrical  Engineering. 

"  It  will  be  a  great  aid  to  students.  The  notes  of  experiments  and  problems  reveai 
much  original  work,  and  the  book  will  be  sure  to  commend  itself  to  instructors." 

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ARRANGED   AND   EDITED   BY 

EDWARD  L.    NICHOLS. 


COMMENTS. 

The  work  as  a  whole  cannot  be  too  highly  commended.  Its  brief  outlines  of  the 
various  experiments  are  very  satisfactory,  its  descriptions  of  apparatus  are  excellent ; 
its  numerous  suggestions  are  calculated  to  develop  the  thinking  and  reasoning  powers 
of  the  student.  The  diagrams  are  carefully  prepared,  and  its  frequent  citations  of 
original  sources  of  information  are  of  the  greatest  value.  —  Street  Railway  Journal. 

The  work  is  clearly  and  concisely  written,  the  fact  that  it  is  edited  by  Professor 
Nichols  being  a  sufficient  guarantee  of  merit.  —  Electrical  Engineering. 

It  will  be  a  great  aid  to  students.  The  notes  of  experiments  and  problems  reveal 
much  original  work,  and  the  book  will  be  sure  to  commend  itself  to  instructors. 

—  S.  F.  Chronicle. 


Immediately  upon  its  publication,  NICHOLS'  LABORATORY  MANUAL  OF 
PHYSICS  AND  APPLIED  ELECTRICITY  became  the  required  text-book  in 
the  following  colleges,  among  others  :  Cornell  University ;  Princeton 
College ;  University  of  Wisconsin ;  University  of  Illinois ;  Tulane 
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versity ;  University  of  Nebraska ;  Brooklyn  Polytechnic  Institute ; 
Maine  State  College ;  Hamilton  College,  Clinton,  N.Y. ;  Wellesley 
College  ;  Mt.  Holyoke  College ;  etc.,  etc. 

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arrangement  of  the  courses  in  Physics  does  not  permit  its  formal 
introduction. 


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CA*.TFO*NH.  LOS  ANGELES 


I 


3  1158  00433  7431 


001264424    1 


